Física 1 - Mecânica 14 Ed. - Resoluções - Capítulo 3

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3-1 
 3.1. IDENTIFY and SET UP: Use =
−
G
G G
2 1
av
2 1
t t
v
r – r
 in component form. 
EXECUTE: (a) 2 1
av-
2 1
5.3 m 1.1 m 1.4 m/s
3.0 s 0x
x x xv
t t t
∆ − −= = = =
∆ − −
 
2 1
av-
2 1
0.5 m 3.4 m 1.3 m/s
3.0 s 0y
y y yv
t t t
∆ − − −= = = = −
∆ − −
 
(b) 
av
av
( ) 1 3 m/stan 0 9286
( ) 1 4 m/s
y
x
v
v
α − .= = = − .
.
 
360 42 9 317α = ° − . ° = ° 
2 2
av av av( ) ( )x yv v v= + 
2 2
av (1 4 m/s) ( 1 3 m/s) 1 9 m/sv = . + − . = . 
Figure 3.1 
 
 
EVALUATE: Our calculation gives that av
Gv is in the 4th quadrant. This corresponds to increasing x and 
decreasing y. 
 3.2. IDENTIFY: Use =
−
G
G G
2 1
av
2 1
t t
v
r – r
 in component form. The distance from the origin is the magnitude of .Gr 
SET UP: At time 1,t 1 1 0.x y= = 
EXECUTE: (a) av-( ) ( 3.8 m/s)(12.0 s) 45.6 mxx v t= ∆ = − = − and av-( ) (4.9 m/s)(12.0 s) 58.8 m.yy v t= ∆ = = 
(b) 2 2 2 2( 45 6 m) (58 8 m) 74 4 m.r x y= + = − . + . = . 
EVALUATE: ∆Gr is in the direction of av.Gv Therefore, x∆ is negative since av-xv is negative and y∆ is 
positive since av-yv is positive. 
 3.3. (a) IDENTIFY and SET UP: From Gr we can calculate x and y for any t. 
Then use =
−
G
G G
2 1
av
2 1t t
v
r – r
 in component form. 
EXECUTE: 2 2 ˆ ˆ[4.0 cm (2.5 cm/s ) ] (5.0 cm/s)t t= + +Gr i j 
At 0,t = ˆ(4 0 cm)= . .Gr i 
At 2 0 s,t = . ˆ ˆ(14 0 cm) (10 0 cm) .= . + .Gr i j 
MOTION IN TWO OR THREE DIMENSIONS 
3
 
3-2 Chapter 3 
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av-
10.0 cm 5.0 cm/s.
2.0 sx
xv
t
∆= = =
∆
 
av-
10.0 cm 5.0 cm/s.
2.0 sy
yv
t
∆= = =
∆
 
 
 2 2
av av av( ) ( ) 7 1 cm/sx yv v v= + = . 
av
av
( )
tan 1 00
( )
y
x
v
v
α = = . 
45 .θ = ° 
Figure 3.3a 
 
EVALUATE: Both x and y increase, so av
Gv is in the 1st quadrant. 
(b) IDENTIFY and SET UP: Calculate Gr by taking the time derivative of ( ).tGr 
EXECUTE: 2 ˆ ˆ([5.0 cm/s ] ) (5.0 cm/s)d t
dt
= = +
GG rv i j 
0:t = 0,xv = 5 0 cm/s;yv = . 5 0 cm/sv = . and 90θ = ° 
1 0 s:t = . 5 0 cm/s,xv = . 5 0 cm/s;yv = . 7 1 cm/sv = . and 45θ = ° 
2 0 s:t = . 10 0 cm/s,xv = . 5 0 cm/s;yv = . 11 cm/sv = and 27θ = ° 
(c) The trajectory is a graph of y versus x. 
2 24 0 cm (2 5 cm/s ) ,x t= . + . (5.0 cm/s)y t= 
For values of t between 0 and 2.0 s, calculate x and y and plot y versus x. 
 
 
Figure 3.3b 
 
EVALUATE: The sketch shows that the instantaneous velocity at any t is tangent to the trajectory. 
 3.4. IDENTIFY: Given the position vector of a squirrel, find its velocity components in general, and at a 
specific time find its velocity components and the magnitude and direction of its position vector and 
velocity. 
Motion in Two or Three Dimensions 3-3 
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SET UP: vx = dx/dt and vy = dy/dt; the magnitude of a vector is 2 2( ).x yA A A= + 
EXECUTE: (a) Taking the derivatives gives 2( ) 0 280 m/s (0 0720 m/s )xv t t= . + . and 
3 2( ) (0.0570 m/s ) .yv t t= 
(b) Evaluating the position vector at 5 00 st = . gives 2 30 mx = . and 2 375 m,y = . which gives 
3 31 m.r = . 
(c) At 5 00 s,t = . 0 64 m/s,xv = + . 1 425 m/s,yv = . which gives 1 56 m/sv = . and 1 425tan
0 64
θ .=
.
 so the 
direction is o65 8θ = . (counterclockwise from +x-axis) 
EVALUATE: The acceleration is not constant, so we cannot use the standard kinematics formulas. 
 3.5. IDENTIFY and SET UP: Use Eq. =
−
G G
G 2 1
av
2 1t t
v – v
a in component form to calculate av-xa and av- .ya 
EXECUTE: (a) The velocity vectors at 1 0t = and 2 30 0 st = . are shown in Figure 3.5a. 
 
 
Figure 3.5a 
 
(b) 22 1
av-
2 1
170 m/s 90 m/s 8 67 m/s
30 0 s
x x x
x
v v va
t t t
∆ − − −= = = = − .
∆ − .
 
2 1 2
av-
2 1
40 m/s 110 m/s 2.33 m/s
30.0 s
y y y
y
v v v
a
t t t
∆ − −= = = = −
∆ −
 
 
(c) 2 2 2
av- av-( ) ( ) 8 98 m/sx ya a a= + = . 
2
av-
2
av-
2.33 m/stan 0.269
8.67 m/s
y
x
a
a
α −= = =
−
 
15 180 195α = ° + ° = ° 
Figure 3.5b 
 
EVALUATE: The changes in xv and yv are both in the negative x or y direction, so both components of 
av
Ga are in the 3rd quadrant. 
 3.6. IDENTIFY: Use =
−
G G
G 2 1
av
2 1
t t
v – v
a in component form. 
SET UP: 2 2 2 2(0.45 m/s )cos31.0 0.39 m/s , (0.45 m/s )sin31.0 0.23 m/sx ya a= ° = = ° = 
EXECUTE: (a) av-
x
x
va
t
∆=
∆
 and 22.6 m/s (0.39 m/s )(10.0 s) 6.5 m/s.xv = + = av-
y
y
v
a
t
∆
=
∆
 and 
21.8 m/s (0.23 m/s )(10.0 s) 0.52 m/s.yv = − + = 
3-4 Chapter 3 
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(b) 2 2(6 5 m/s) (0 52 m/s) 6 52 m/s,v = . + . = . at an angle of 0.52arctan 4.6
6.5
⎛ ⎞ = °⎜ ⎟
⎝ ⎠
 counterclockwise from 
the +x-axis. 
(c) The velocity vectors 1
Gv and 2
Gv are sketched in Figure 3.6. The two velocity vectors differ in 
magnitude and direction. 
EVALUATE: 1
Gv is at an angle of 35° below the +x-axis and has magnitude 1 3 2 m/s,v = . so 2 1v v> and 
the direction of 2
Gv is rotated counterclockwise from the direction of 1.Gv 
 
 
Figure 3.6 
 
 3.7. IDENTIFY and SET UP: Use d
dt
=
GG rv and =
G
G
d
dt
a
v
 to find ,xv ,yv ,xa and ya as functions of time. The 
magnitude and direction of Gr and Ga can be found once we know their components. 
EXECUTE: (a) Calculate x and y for t values in the range 0 to 2.0 s and plot y versus x. The results are 
given in Figure 3.7a. 
 
 
Figure 3.7a 
 
(b) x
dxv
dt
α= = 2y
dyv t
dt
β= = − 
0x
x
dva
dt
= = 2y
y
dv
a
dt
β= = − 
Thus ˆ ˆ2 tα β= −Gv i j , ˆ2β= −Ga j 
(c) velocity: At 2 0 s,t = . 2 4 m/s,xv = . 22(1.2 m/s )(2.0 s) 4.8 m/syv = − = − 
 
Motion in Two or Three Dimensions 3-5 
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2 2 5 4 m/sx yv v v= + = . 
4 8 m/stan 2 00
2 4 m/s
y
x
v
v
α − .= = = − .
.
 
63 4 360 297α = − . ° + ° = ° 
Figure 3.7b 
 
acceleration: At 2 0 s,t = . 0,xa = 2 22 (1 2 m/s ) 2 4 m/sya = − . = − . 
 
 
2 2 22 4 m/sx ya a a= + = . 
22.4 m/stan
0
y
x
a
a
β −= = = −∞ 
270β = ° 
Figure 3.7c 
 
 
EVALUATE: (d) Ga has a component a& in the same 
direction as ,Gv so we know that v is increasing (the bird 
is speeding up). Ga also has a component a⊥ 
perpendicular to ,Gv so that the direction of Gv is 
changing; the bird is turning toward the -directiony− 
(toward the right) 
Figure 3.7d 
 
Gv is always tangent to the path; Gv at 2 0 st = . shown in part (c) is tangent to the path at this t, conforming 
to this general rule. Ga is constant and in the -direction;y− the direction of Gv is turning toward the 
-directiony− . 
 3.8. IDENTIFY: Use the velocity components of a car (given as a function of time) to find the acceleration of 
the car as a function of time and to find the magnitude and direction of the car’s velocity and acceleration 
at axv and .yv 
EXECUTE: Combining Eqs. 3.24, 3.21 and 3.22 gives 
2 2 2 2 2 2 2 2
0 0 0 0 0 0 0 0 0cos ( sin ) (sin cos ) 2 sin ( ) .v v v gt v v gt gtα α α α α= + − = + − + 
Motion in Two or Three Dimensions 3-29 
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2 2 2 2
0 0 0 0
12 sin 2 ,
2
v v g v t gt v gyα⎛ ⎞= − − = −⎜ ⎟
⎝ ⎠
 where Eq. (3.20) has been used to eliminate t in favor of y. For 
the case of a rock thrown from the roof of a building of height h, the speed at the ground is found by 
substituting y h= − into the above expression, yielding 2
0 2 ,v v gh= + which is independent of 0.α 
EVALUATE: This result, as will be seen in the chapter dealing with conservation of energy (Chapter 7), is 
valid for any y, positive, negative or zero, as long as 2
0 2 0.v gy− > 
 3.63. (a) IDENTIFY: Projectile motion. 
 
 Take the origin of coordinates at the top of 
the ramp and take y+ to be upward. 
The problem specifies that the object is 
displaced 40.0 m to the right when it is 
15.0 m below the origin. 
 
Figure 3.63 
 
We don’t know t, the time in the air, and we don’t know 0v . Write down the equations for the horizontal 
and vertical displacements. Combine these two equations to eliminate one unknown. 
SET UP: y-component: 
0 15 0 m,y y− = − . 29 80 m/s ,ya = − . 0 0 sin53 0yv v= . ° 
21
0 0 2y yy y v t a t− = + 
EXECUTE: 2 2
015 0 m ( sin53 0 ) (4 90 m/s )v t t− . = . ° − . 
SET UP: x-component: 
0 40 0 m,x x− = . 0,xa = 0 0 cos53 0xv v= . ° 
21
0 0 2x xx x v t a t− = + 
EXECUTE: 040 0 m ( )cos53 0v t. = . ° 
The second equation says 0
40 0 m 66 47 m
cos53 0
v t .= = . .
. °
 
Use this to replace 0v t in the first equation: 
2 215 0 m (66 47 m) sin53 (4 90 m/s ) t− . = . ° − . 
2 2
(66 47 m)sin53 15 0 m 68 08 m 3 727 s
4 90 m/s 4 90 m/s
t . ° + . .= = = . .
. .
 
Now that we have t we can use the x-component equation to solve for 0:v 
0
40 0 m 40 0 m 17 8 m/s
cos53 0 (3 727 s) cos53 0
v
t
. .= = = . .
. ° . . °
 
EVALUATE: Using these values of 0v and t in the 21
0 0 2y yy y v a t= = + equation verifies that 
0 15 0 my y− = − . . 
(b) IDENTIFY: 0 (17 8 m/s)/2 8 9 m/sv = . = . 
This is less than the speed required to make it to the other side, so he lands in the river. 
Use the vertical motion to find the time it takes him to reach the water: 
SET UP: 0 100 m;y y− = − 0 0 sin53.0 7.11 m/s;yv v= + ° = 29 80 m/sya = − . 
21
0 0 2y yy y v t a t− = + gives 2100 7 11 4 90t t− = . − . 
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EXECUTE: 24 90 7 11 100 0t t. − . − = and ( )21
9 80 7 11 (7 11) 4 (4 90)( 100)t .= . ± . − . − 
0 726 s 4 57 st = . ± . so 5 30 st = . . 
The horizontal distance he travels in this time is 
0 0 0( cos53 0 ) (5 36 m/s)(5 30 s) 28 4 mxx x v t v t− = = . ° = . . = . . 
He lands in the river a horizontal distance of 28.4 m from his launch point. 
EVALUATE: He has half the minimum speed and makes it only about halfway across. 
 3.64. IDENTIFY: The ball moves in projectile motion. 
SET UP: The woman and ball travel for the same time and must travel the same horizontal distance, so for 
the ball 0 6 00 m/s.xv = . 
EXECUTE: (a) 0 0 0cos .xv v θ= 0
0
0
6.00 m/scos
20.0 m/s
xv
v
θ = = and 0 72 5 .θ = . ° The ball is in the air for 5.55s and 
she runs a distance of (6.00 m/s)(5.55 s) = 33.3 m. 
(b) Relative to the ground the ball moves in a parabola. The ball and the runner have the same horizontal 
component of velocity, so relative to the runner the ball has only vertical motion. The trajectories as seen 
by each observer are sketched in Figure 3.64. 
EVALUATE: The ball could be thrown with a different speed, so long as the angle at which it was thrown 
was adjusted to keep 0 6 00 m/s.xv = . 
 
 
Figure 3.64 
 
 3.65. IDENTIFY: The boulder moves in projectile motion. 
SET UP: Take y+ downward. 0 0 ,xv v= 0,xa = 0,xa = 29 80 m/s .ya = + . 
EXECUTE: (a) Use the vertical motion to find the time for the boulder to reach the level of the lake: 
21
0 0 2y yy y v t a t− = + with 0 20 my y− = + gives 0
2
2( ) 2(20 m) 2 02 s.
9 80 m/sy
y yt
a
−= = = .
.
 The rock must 
travel horizontally 100 m during this time. 21
0 0 2x xx x v t a t− = + gives 
0
0 0
100 m 49 5 m/s
2 02 sx
x xv v
t
−= = = = .
.
 
(b) In going from the edge of the cliff to the plain, the boulder travels downward a distance of 
0 45 m.y y− = 0
2
2( ) 2(45 m) 3 03 s
9 80 m/sy
y yt
a
−= = = .
.
 and 0 0 (49 5 m/s)(3 03 s) 150 m.xx x v t− = = . . = 
The rock lands 150 m 100 m 50 m− = beyond the foot of the dam. 
EVALUATE: The boulder passes over the dam 2.02 s after it leaves the cliff and then travels an additional 1.01 s 
before landing on the plain. If the boulder has an initial speed that is less than 49 m/s, then it lands in the lake. 
 3.66. IDENTIFY: The bagels move in projectile motion. Find Henrietta’s location when the bagels reach the 
ground, and require the bagels to have this horizontal range. 
SET UP: Let y+ be downward and let 0 0 0.x y= = 0,xa = .ya g= + When the bagels reach the ground, 
38 0 m.y = . 
EXECUTE: (a) When she catches the bagels, Henrietta has been jogging for 9.00 s plus the time for the 
bagels to fall 38.0 m from rest. Get the time to fall: 21 ,
2
y gt= 2 2138 0 m (9 80 m/s )
2
t. = . and 2 78 s.t = . 
So, she has been jogging for 9 00 s 2 78 s 11 78 s.. + . = . During this time she has gone 
Motion in Two or Three Dimensions 3-31 
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(3 05 m/s)(11 78 s) 35 9 m.x vt= = . . = . Bruce must throw the bagels so they travel 35.9 m horizontally in 
2.78 s. This gives .x vt= 35 9 m (2 78 s)v. = . and 12 9 m/s.v = . 
(b) 35.9 m from the building. 
EVALUATE: If 12 9 m/sv > . the bagels land in front of her and if 12 9 m/sv0xa = and 29 80 m/sya = − . . The water has 0 0 0cos 15 0 m/sxυ υ θ= = . and 0 0 0sin 20 0 m/syυ υ θ= = . . 
EXECUTE: Use the vertical motion to find t that gives 0 10 0 m:y y− = . 21
0 0 2y yy y t a tυ− = + gives 
2 210 0 m (20 0 m/s) (4 90 m/s )t t. = . − . . 
The quadratic formula gives 2 04 1 45 s,t = . ± . and 0 59 st = . or 3 49 st = . . Both answers are physical. 
For 0 59 s,t = . 0 0 (15 0 m/s)(0 59 s) 8 8 mxx x tυ− = = . . = . . 
For 3 49 s,t = . 0 0 (15 0 m/s)(3 49 s) 52 4 mxx x tυ− = = . . = . . 
When the cannon is 8.8 m from the building, the water hits this spot on the wall on its way up to its 
maximum height. When is it 52.4 m from the building it hits this spot after it has passed through its 
maximum height. 
EVALUATE: The fact that we have two possible answers means that the firefighters have some choice on 
where to stand. If the fire is extremely fierce, they would no doubt prefer to stand at the more distant 
location. 
 3.69. IDENTIFY: The rock is in free fall once it is in the air, so it has only a downward acceleration of 9.80 m/s2, 
and we apply the principles of two-dimensional projectile motion to it. The constant-acceleration 
kinematics formulas apply. 
SET UP: The vertical displacement must be ∆ = − 0y y y = 5.00 m – 1.60 m = 3.40 m at the instant that 
the horizontal displacement ∆ = − 0x x x = 14.0 m, and ay = –9.80 m/s2 with +y upward. 
EXECUTE: (a) There is no horizontal acceleration, so 14.0 m = v0 cos(56.0°)t, which gives 
=
°0
14.0 m
.
cos 56.0
t
v
 Putting this quantity, along with the numerical quantities, into the equation 
21
0 0 2y yy y v t a t− = + and solving for v0 we get v0 = 13.3 m/s. 
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(b) The initial horizontal velocity of the rock is (13.3 m/s)(cos 56.0°), and when it lands on the ground, 
− 0y y = –1.60 m. Putting these quantities into the equation 21
0 0 2y yy y v t a t− = + leads to a quadratic 
equation in t. Using the positive square root, we get t = 2.388 s when the rock lands. The horizontal 
position at that instant is − 0x x = (13.3 m/s)(cos 56.0°)(2.388 s) = 17.8 m from the launch point. So the 
distance beyond the fence is 17.8 m – 14.0 m = 3.8 m. 
EVALUATE: We cannot use the range formula to find the distance in (b) because the rock’s motion does 
not start and end at the same height. 
 3.70. IDENTIFY: The object moves with constant acceleration in both the horizontal and vertical directions. 
SET UP: Let y+ be downward and let x+ be the direction in which the firecracker is thrown. 
EXECUTE: The firecracker’s falling time can be found from the vertical motion: 2 .ht
g
= 
The firecracker’s horizontal position at any time t (taking the student’s position as 0x = ) is 21
2 .x vt at= − 
0x = when cracker hits the ground, so 2 .t v/a= Combining this with the expression for the falling time 
gives 2 2v h
a g
= and 
2
2
2 .v gh
a
= 
EVALUATE: When h is smaller, the time in the air is smaller and either v must be smaller or a must be 
larger. 
 3.71. IDENTIFY: Relative velocity problem. The plane’s motion relative to the earth is determined by its 
velocity relative to the earth. 
SET UP: Select a coordinate system where y+ is north and x+ is east. 
The velocity vectors in the problem are: 
P/E ,Gv the velocity of the plane relative to the earth. 
P/A ,Gv the velocity of the plane relative to the air (the magnitude P/Av is the airspeed of the plane and the 
direction of P/A
Gv is the compass course set by the pilot). 
A/E ,Gv the velocity of the air relative to the earth (the wind velocity). 
The rule for combining relative velocities gives P/E P/A A/E= + .G G Gv v v 
 (a) We are given the following information about the relative velocities: 
P/A
Gv has magnitude 220 km/h and its direction is west. In our coordinates it has components 
P/A( ) 220 km/hxv = − and P/A( ) 0yv = . 
From the displacement of the plane relative to the earth after 0.500 h, we find that P/E
Gv has components in 
our coordinate system of 
P/E
120 km( ) 240 km/h
0 500 hxv = − = −
.
 (west) 
P/E
20 km( ) 40 km/h
0 500 hyv = − = −
.
 (south) 
With this information the diagram corresponding to the velocity addition equation is shown in 
Figure 3.71a. 
 
 
Figure 3.71a 
 
 
 
 
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We are asked to find A/E ,Gv so solve for this vector: 
P/E P/A A/E= +G G Gv v v gives A/E P/E P/A= − .G G Gv v v 
EXECUTE: The x-component of this equation gives 
A/E P/E P/A( ) ( ) ( ) 240 km/h ( 220 km/h) 20 km/hx x xv v v= − = − − − = − . 
The y-component of this equation gives 
A/E P/E P/A( ) ( ) ( ) 40 km/hy y yv v v= − = − . 
Now that we have the components of A/E
Gv we can find its magnitude and direction. 
 
 2 2
A/E A/E A/E( ) ( )x yv v v= + 
2 2
A/E ( 20 km/h) ( 40 km/h) 44 7 km/hv = − + − = . 
40 km/htan 2 00;
20 km/h
φ = = . 63 4φ = . ° 
The direction of the wind velocity is 63 4 S. ° of W, 
or 26 6 W. ° of S. 
Figure 3.71b 
 
EVALUATE: The plane heads west. It goes farther west than it would without wind and also travels south, 
so the wind velocity has components west and south. 
(b) SET UP: The rule for combining the relative velocities is still P/E P/A A/E ,= +G G Gv v v but some of these 
velocities have different values than in part (a). 
P/A
Gv has magnitude 220 km/h but its direction is to be found. 
A/E
Gv has magnitude 40 km/h and its direction is due south. 
The direction of P/E
Gv is west; its magnitude is not given. 
The vector diagram for P/E P/A A/E= +G G Gv v v and the specified directions for the vectors is shown in 
Figure 3.71c. 
 
 
Figure 3.71c 
 
The vector addition diagram forms a right triangle. 
EXECUTE: A/E
P/A
40 km/hsin 0 1818;
220 km/h
v
v
φ = = = . 10 5φ = . °. 
The pilot should set her course 10 5. ° north of west. 
EVALUATE: The velocity of the plane relative to the air must have a northward component to counteract 
the wind and a westward component in order to travel west. 
 
 
 
 
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 3.72. IDENTIFY: Use the relation that relates the relative velocities. 
SET UP: The relative velocities are the raindrop relative to the earth, R/E ,Gv the raindrop relative to the 
train, R/T ,Gv and the train relative to the earth, T/E.Gv R/E R/T T/E.= +G G Gv v v T/E
Gv is due east and has 
magnitude 12.0 m/s. R/T
Gv is 30 0. ° west of vertical. R/E
Gv is vertical. The relative velocity addition 
diagram is given in Figure 3.72. 
EXECUTE: (a) R/E
Gv is vertical and has zero horizontal component. The horizontal component of R/T
Gv is 
T/E ,−Gv so is 12.0 m/s westward. 
(b) T/E
R/E
12 0 m/s 20 8 m/s.
tan30 0 tan30 0
vv .= = = .
. ° . °
 T/E
R/T
12 0 m/s 24 0 m/s.
sin30 0 sin30 0
vv .= = = .
. ° . °
 
EVALUATE: The speed of the raindrop relative to the train is greater than its speed relative to the earth, 
because of the motion of the train. 
 
 
Figure 3.72 
 
 3.73. IDENTIFY: Relative velocity problem. 
SET UP: The three relative velocities are: 
J/G ,Gv Juan relative to the ground. This velocity is due north and has magnitude J/G 8.00 m/s.v = 
B/G ,Gv the ball relative to theground. This vector is 37 0. ° east of north and has magnitude 
B/G 12 00 m/sv = . . 
B/J ,Gv the ball relative to Juan. We are asked to find the magnitude and direction of this vector. 
The relative velocity addition equation is B/G B/J J/G ,= +G G Gv v v so B/J B/G J/G= − .G G Gv v v 
The relative velocity addition diagram does not form a right triangle so we must do the vector addition 
using components. 
Take y+ to be north and x+ to be east. 
EXECUTE: B/J B/G sin37 0 7 222 m/sxv v= + . ° = . 
B/J B/G J/Gcos37 0 1 584 m/syv v v= + . ° − = . 
These two components give B/J 7 39 m/sv = . at 12 4. ° north of east. 
EVALUATE: Since Juan is running due north, the ball’s eastward component of velocity relative to him is 
the same as its eastward component relative to the earth. The northward component of velocity for Juan 
and the ball are in the same direction, so the component for the ball relative to Juan is the difference in 
their components of velocity relative to the ground. 
 3.74. IDENTIFY: Both the bolt and the elevator move vertically with constant acceleration. 
SET UP: Let y+ be upward and let 0y = at the initial position of the floor of the elevator, so 0y for the 
bolt is 3.00 m. 
EXECUTE: (a) The position of the bolt is 2 23 00 m (2 50 m/s) (1/ 2)(9 80 m/s )t t. + . − . and the position of 
the floor is (2.50 m/s)t. Equating the two, 2 23 00 m (4 90 m/s ) .t. = . Therefore, 0 782 s.t = . 
(b) The velocity of the bolt is 22 50 m/s (9 80 m/s )(0 782 s) 5 17 m/s. − . . = − . relative to earth, therefore, 
relative to an observer in the elevator 5 17 m/s 2 50 m/s 7 67 m/sv = − . − . = − . . 
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(c) As calculated in part (b), the speed relative to earth is 5.17 m/s. 
(d) Relative to earth, the distance the bolt traveled is 
2 2 2 2(2 50 m/s) (1/ 2)(9 80 m/s ) (2 50 m/s)(0 782 s) (4 90 m/s )(0 782 s) 1 04 m.t t. − . = . . − . . = − . 
EVALUATE: As viewed by an observer in the elevator, the bolt has 0 0yv = and 29 80 m/s ,ya = − . so in 
0.782 s it falls 2 21
2 (9 80 m/s )(0 782 s) 3 00 m.− . . = − . 
3.75. IDENTIFY: We need to use relative velocities. 
SET UP: If B is moving relative to M and M is moving relative to E, the velocity of B relative to E is 
B/E B/M M/E.= +G G Gv v v 
EXECUTE: Let +x be east and +y be north. We have B/M, 2 50 m/s,xv = . B/M,y 4 33 m/s,v = − . M/E, 0,xv = 
and M/E,y 6 00 m/s.v = . Therefore B/E, B/M, M/E, 2 50 m/sx x xv v v= + = . and 
B/E,y B/M,y M/E,y 4 33 m/s 6 00 m/s 1 67 m/s.v v v= + = − . + . = + . The magnitude is 
2 2
B/E (2 50 m/s) (1 67 m/s) 3 01 m/s,v = . + . = . and the direction is 1 67tan ,
2 50
θ .=
.
 which gives 
o33 7θ = . north of east. 
EVALUATE: Since Mia is moving, the velocity of the ball relative to her is different from its velocity 
relative to the ground or relative to Alice. 
 3.76. IDENTIFY: You have a graph showing the horizontal range of the rock as a function of the angle at which 
it was launched and want to find its initial velocity. Because air resistance is negligible, the rock is in free 
fall. The range formula applies since the rock rock was launced from the ground and lands at the ground. 
SET UP: (a) The range formula is 
θ
=
2
0 sin(2 )v
R
g
, so a plot of R versus θ0sin(2 ) will give a straight line 
having slope equal to 2
0
/ .v g We can use that data in the graph in the problem to construct our graph by 
hand, or we can use graphing software. The resulting graph is shown in Figure 3.76. 
 
 
Figure 3.76 
 
(b) The slope of the graph is 10.95 m, so 10.95 m = 2
0 / .v g Solving for v0 we get v0 = 10.4 m/s. 
(c) Solving the formula = + −2 2
0 02 ( )y y yv v a y y for − 0y y with vy = 0 at the highest point, we get 
y – y0 = 1.99 m. 
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EVALUATE: This approach to finding the launch speed v0 requires only simple measurements: the range 
and the launch angle. It would be difficult and would require special equipment to measure v0 directly. 
 3.77. IDENTIFY: The table gives data showing the horizontal range of the potato for various launch heights. 
You want to use this information to determine the launch speed of the potato, assuming negligible air 
resistance. 
SET UP: The potatoes are launched horizontally, so v0y = 0, and they are in free fall, so ay = 9.80 m/s2 
downward and ax = 0. The time a potato is in the air is just the time it takes for it to fall vertically from the 
launch point to the ground, a distance h. 
EXECUTE: (a) For the vertical motion of a potato, we have h = ½ gt2, so t = 2 / .h g The horizontal range 
R is given by 0 0 2 / .R v t v h g= = Squaring gives =
⎛ ⎞
⎜ ⎟
⎝ ⎠
2
2 02
.
v
R h
g
 Graphing R2 versus h will give a straight 
line with slope 2
02 / .v g We can graph the data from the table in the text by hand, or we could use graphing 
software. The result is shown in Figure 3.77. 
 
 
Figure 3.77 
 
(b) The slope of the graph is 55.2 m, so =
2
0
(9.80 m/s )(55.2 m)
2
v = 16.4 m/s. 
(c) In this case, the potatoes are launched and land at ground level, so we can use the range formula with θ 
= 30.0° and v0 = 16.4 m/s. The result is 
2
0 sin(2 )vR
g
θ= = 23.8 m. 
EVALUATE: This approach to finding the launch speed v0 requires only simple measurements: the range 
and the launch height. It would be difficult and would require special equipment to measure v0 directly. 
 3.78. IDENTIFY: This is a vector addition problem. The boat moves relative to the water and the water moves 
relative to the earth. We know the speed of the boat relative to the water and the times for the boat to go 
directly across the river, and from these things we want to find out how fast the water is moving and the 
width of the river. 
SET UP: For both trips of the boat, = +G G G
B/E B/W W/Ev v v , where the subscripts refer to the boat, earth, and 
water. The speed of the boat relative to the earth is vB/E = d/t, where d is the width of the river and t is the 
time to cross the river, which is different in the two crossings. 
EXECUTE: Figure 3.78 shows a vector sum for the first trip and for the return trip. 
 
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Figure 3.78a-b 
 
(a) For both trips, the vectors in Figures 3.78 a & b form right triangles, so we can apply the Pythagorean 
theorem. = −2 2 2
B/E B/W W/Ev v v and vB/E = d/t. For the first trip, vB/W = 6.00 m/s and t = 20.1 s, giving 
2 2 2 2
W/E/(20.1s) (60.00 m/s ) ( ) .d v= − For the return trip, vB/W = 9.0 m/s and t = 11.2 s, which gives 
2 2 2 2
W/E/(11.2 s) (9.00 m/s ) ( ) .d v= − Solving these two equations together gives d = 90.48 m, which 
rounds to 90.5 m (the width of the river) and vW/E = 3.967 m/s which rounds to 3.97 m/s (the speed of the 
current). 
(b) The shortest time is when the boat heads perpendicular to the current, which is due north. Figure 3.78c 
illustrates this situation. The time to cross is t = d/vB/W = (90.48 m)/(6.00 m/s) = 15.1 s. The distance x east 
(down river) that you travel is x = vW/Et = (3.967 m/s)(15.1 s) = 59.9 m east of your starting point. 
 
 
Figure 3.78c 
 
EVALUATE: In part (a), the boat must havea velocity component up river to cancel out the current 
velocity. In part (b), velocity of the current has no effect on the crossing time, but it does affect the landing 
position of the boat. 
 3.79. IDENTIFY: Write an expression for the square of the distance 2( )D from the origin to the particle, 
expressed as a function of time. Then take the derivative of 2D with respect to t, and solve for the value 
of t when this derivative is zero. If the discriminant is zero or negative, the distance D will never decrease. 
SET UP: 2 2 2 ,D x y= + with ( )x t and ( )y t given by Eqs. (3.19) and (3.20). 
EXECUTE: Following this process, 1sin 8/9 70 5− = . °. 
EVALUATE: We know that if the object is thrown straight up it moves away from P and then returns, so 
we are not surprised that the projectile angle must be less than some maximum value for the distance to 
always increase with time. 
 3.80. IDENTIFY: Apply the relative velocity relation. 
SET UP: Let C/Wv be the speed of the canoe relative to water and W/Gv be the speed of the water relative 
to the ground. 
EXECUTE: (a) Taking all units to be in km and h, we have three equations. We know that heading 
upstream C/W W/G 2.v v− = We know that heading downstream for a time C/W W/G, ( ) 5.t v v t+ = We also 
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know that for the bottle W/G ( 1) 3.v t + = Solving these three equations for W/G C/W, 2 ,v x v x= = + 
therefore (2 ) 5x x t+ + = or (2 2 ) 5.x t+ = Also 3/ 1,t x= − so 3(2 2 ) 1 5x
x
⎛ ⎞+ − =⎜ ⎟
⎝ ⎠
 or 22 6 0.x x+ − = 
The positive solution is W/G 1.5 km/h.x v= = 
(b) C/W W/G2 km/h 3.5 km/h.v v= + = 
EVALUATE: When they head upstream, their speed relative to the ground is 
3.5 km/h 1.5 km/h 2.0 km/h.− = When they head downstream, their speed relative to the ground is 
3.5 km/h 1.5 km/h 5.0 km/h.+ = The bottle is moving downstream at 1.5 km/s relative to the earth, so they 
are able to overtake it. 
 3.81. IDENTIFY: The rocket has two periods of constant acceleration motion. 
SET UP: Let y+ be upward. During the free-fall phase, 0xa = and .ya g= − After the engines turn on, 
(3 00 )cos30 0xa g= . . ° and (3 00 )sin30 0 .ya g= . . ° Let t be the total time since the rocket was dropped and 
let T be the time the rocket falls before the engine starts. 
EXECUTE: (i) The diagram is given in Figure 3.81 a. 
(ii) The x-position of the plane is (236 m/s)t and the x-position of the rocket is 
2 2(236 m/s) (1 2)(3.00)(9.80 m/s )cos30 ( ) .t / t T+ ° − The graphs of these two equations are sketched in 
Figure 3.81 b. 
(iii) If we take 0y = to be the altitude of the airliner, then 
2 2 2( ) 1 2 ( ) 1 2(3.00)(9.80 m/s )(sin30 )( )y t / gT gT t T / t T= − − − + ° − for the rocket. The airliner has constant y. 
The graphs are sketched in Figure 3.81b. 
In each of the Figures 3.81a–c, the rocket is dropped at 0t = and the time T when the motor is turned on is 
indicated. 
By setting 0y = for the rocket, we can solve for t in terms of T: 
2 2 2 2 20 4.90 m/s ) (9.80 m/s ) ( ) (7.35 m/s )( ) .T T t T t T= − − − + −( Using the quadratic formula for the 
variable x t T= − we find 
2 2 2 2 2
2
(9.80 m/s ) (9.80 m/s ) (4)(7.35 m/s )(4.9)
,
2(7.35 m/s )
T T T
x t T
+ +
= − = or 
2 72 .t T= . Now, using the condition that rocket plane 1000 m,x x− = we find 
2 2(236 m/s) (12.7 m/s )( ) (236 m/s) 1000 m,t t T t + − − = or 2 2(1 72 ) 78 6 s .T. = . Therefore 5 15 s.T = . 
EVALUATE: During the free-fall phase the rocket and airliner have the same x coordinate but the rocket 
moves downward from the airliner. After the engines fire, the rocket starts to move upward and its 
horizontal component of velocity starts to exceed that of the airliner. 
 
Figure 3. 81 
 
 
 3.82. IDENTIFY: We know the speed of the seeds and the distance they travel. 
SET UP: We can treat the speed as constant over a very short distance, so v = d/t. The minimum frame 
rate is determined by the maximum speed of the seeds, so we use v = 4.6 m/s. 
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EXECUTE: Solving for t gives t =d/v = (0.20 × 10–3 s)/(4.6 m/s) = 4.3 × 10–5 s per frame. 
The frame rate is 1/(4.3 × 10–5 s per frame) = 23,000 frames/seconde. Choice (c) 25,000 frames per second 
is closest to this result, so choice (c) is the best one. 
EVALUATE: This experiment would clearly require high-speed photography. 
 3.83. IDENTIFY: A seed launched at 90° goes straight up. Since we are ignoring air resistance, its acceleration is 
9.80 m/s2 downward. 
SET UP: For the highest possible speed v0y = 4.6 m/s, and vy = 0 at the highest point. 
EXECUTE: vy = v0y – gt gives t = v0y/g = (4.6 m/s)/(9.80 m/s2) = 0.47 s, which is choice (b). 
EVALUATE: Seeds are rather light and 4.6 m/s is fairly fast, so it might not be such a good idea to ignore 
air resistance. But doing so is acceptable to get a first approximation to the time. 
 3.84. IDENTIFY: A seed launched at 0° starts out traveling horizontally from a height of 20 cm above the 
ground. Since we are ignoring air resistance, its acceleration is 9.80 m/s2 downward. 
SET UP: Its horizontal distance is determined by the time it takes the seed to fall 20 cm, starting from rest 
vertically. 
EXECUTE: The time to fall 20 cm is 21
20.20 m ,gt= which gives t = 0.202 s. The horizontal distance 
traveled during this time is x = (4.6 m/s)(0.202 s) = 0.93 m = 93 cm, which is choice (b). 
EVALUATE: In reality the seed would travel a bit less distance due to air resistance. 
 3.85. IDENTIFY: About 2/3 of the seeds are launched between 6° and 56° above the horizontal, and the average 
for all the seeds is 31°. So clearly most of the seeds are launched above the horizontal. 
SET UP and EXECUTE: For choice (a) to be correct, the seeds would need to cluster around 90°, which they 
do not. For choice (b), most seeds would need to launch below the horizontal, which is not the case. For 
choice (c), the launch angle should be around +45°. Since 31° is not far from 45°, this is the best choice. 
For choice (d), the seeds should go straight downward. This would require a launch angle of –90°, which is 
not the case. 
EVALUATE: Evolutionarily it would be an advantage for the seeds to get as far from the parent plant as 
possible so the young plants would not compete with the parent for water and soil nutrients, so 45° is a 
biologically plausible result. Natural selection would tend to favor plants that launched their seeds at this 
angle over those that did not.specific time. 
SET UP: /x xa dv dt= and / ;y ya dv dt= the magnitude of a vector is 2 2( ).x yA A A= + 
EXECUTE: (a) Taking the derivatives gives 3( ) ( 0.0360 m/s )xa t t= − and 2( ) 0 550 m/s .ya t = . 
(b) Evaluating the velocity components at 8 00 st = . gives 3 848 m/sxv = . and 6 40 m/s,yv = . which gives 
7 47 m/s.v = . The direction is 6 40tan
3 848
θ .=
.
 so o59 0θ = . (counterclockwise from +x-axis). 
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(c) Evaluating the acceleration components at 8 00 st = . gives 20 288 m/sxa = .2 and 20 550 m/s ,ya = . 
which gives 20 621 m/s .a = . The angle with the +y axis is given by 0 288tan ,
0 550
θ .=
.
 so o27 6 .θ = . The 
direction is therefore o118 counterclockwise from +x-axis. 
EVALUATE: The acceleration is not constant, so we cannot use the standard kinematics formulas. 
 3.9. IDENTIFY: The book moves in projectile motion once it leaves the tabletop. Its initial velocity is 
horizontal. 
SET UP: Take the positive y-direction to be upward. Take the origin of coordinates at the initial position 
of the book, at the point where it leaves the table top. 
 
 
x-component: 
0,xa = 0 1.10 m/s,xv = 
0.480 st = 
y-component: 
29.80 m/s ,ya = − 
0 0,yv = 
0.480 st = 
Figure 3.9a 
 
 
Use constant acceleration equations for the x and y components of the motion, with 0xa = and .ya g= − 
EXECUTE: (a) 0 ?y y− = 
2 2 21 1
0 0 2 20 ( 9.80 m/s )(0.480 s) 1.129 m.y yy y v t a t− = + = + − = − The tabletop is therefore 1.13 m above 
the floor. 
(b) 0 ?x x− = 
21
0 0 2 (1.10 m/s)(0.480 s) 0 0.528 m.x xx x v t a t− = + = + = 
(c) 0 1 10 m/sx x xv v a t= + = . (The x-component of the velocity is constant, since 0.)xa = 
2
0 0 ( 9.80 m/s )(0.480 s) 4.704 m/sy y yv v a t= + = + − = − 
 
2 2 4.83 m/sx yv v v= + = 
4.704 m/stan 4.2764
1 10 m/s
y
x
v
v
α −= = = −
.
 
76.8α = − ° 
Direction of Gv is 76.8° below the horizontal 
Figure 3.9b 
 
(d) The graphs are given in Figure 3.9c. 
 
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Figure 3.9c 
 
EVALUATE: In the x-direction, 0xa = and xv is constant. In the y-direction, 29 80 m/sya = − . and yv is 
downward and increasing in magnitude since ya and yv are in the same directions. The x and y motions 
occur independently, connected only by the time. The time it takes the book to fall 1.13 m is the time it 
travels horizontally. 
 3.10. IDENTIFY: The person moves in projectile motion. She must travel 1.75 m horizontally during the time 
she falls 9.00 m vertically. 
SET UP: Take y+ downward. 0,xa = 29.80 m/s .ya = + 0 0 ,xv v= 0 0.yv = 
EXECUTE: Time to fall 9.00 m: 21
0 0 2y yy y v t a t− = + gives 0
2
2( ) 2(9 00 m) 1 36 s.
9 80 m/sy
y yt
a
− .= = = .
.
 
Speed needed to travel 1.75 m horizontally during this time: 21
0 0 2x xx x v t a t− = + gives 
0
0 0
1 75 m 1 29 m/s.
1 36 sx
x xv v
t
− .= = = = .
.
 
EVALUATE: If she increases her initial speed she still takes 1.36 s to reach the level of the ledge, but has 
traveled horizontally farther than 1.75 m. 
 3.11. IDENTIFY: Each object moves in projectile motion. 
SET UP: Take y+ to be downward. For each cricket, 0xa = and 29 80 m/s .ya = + . For Chirpy, 
0 0 0.x yv v= = For Milada, 0 0 950 m/s,xv = . 0 0.yv = 
EXECUTE: Milada’s horizontal component of velocity has no effect on her vertical motion. She also 
reaches the ground in 2.70 s. 21
0 0 2 (0 950 m/s)(2.70 s) 2.57 m.x xx x v t a t− = + = . = 
EVALUATE: The x and y components of motion are totally separate and are connected only by the fact that 
the time is the same for both. 
 3.12. IDENTIFY: The football moves in projectile motion. 
SET UP: Let y+ be upward. 0,xa = .ya g= − At the highest point in the trajectory, 0.yv = 
EXECUTE: (a) 0 .y y yv v a t= + The time t is 0
2
12 0 m s 1 224 s,
9 80 m/s
yv /
g
. = = .
. 
 which we round to 1.22 s. 
(b) Different constant acceleration equations give different expressions but the same numerical result: 
2
021 1
02 2 7 35 m.
2
y
y
v
gt v t
g
= = = . 
(c) Regardless of how the algebra is done, the time will be twice that found in part (a), which is 
2(1.224 s) = 2.45 s. 
(d) 0,xa = so 0 0 (20 0 m/s)(2 45 s) 49 0 m.xx x v t− = = . . = . 
(e) The graphs are sketched in Figure 3.12 (next page). 
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EVALUATE: When the football returns to its original level, 20 0 m/sxv = . and 12 0 m/s.yv = − . 
 
Figure 3.12 
 
 3.13. IDENTIFY: The car moves in projectile motion. The car travels 21 3 m 1 80 m 19 5 m. − . = . downward 
during the time it travels 48.0 m horizontally. 
SET UP: Take y+ to be downward. 0,xa = 29 80 m/s .ya = + . 0 0 ,xv v= 0 0.yv = 
EXECUTE: (a) Use the vertical motion to find the time in the air: 
21
0 0 2y yy y v t a t− = + gives 0
2
2( ) 2(19 5 m) 1 995 s
9 80 m/sy
y yt
a
− .= = = .
.
 
Then 21
0 0 2x xx x v t a t− = + gives 0
0 0
48 0 m 24.1 m/s.
1 995 sx
x xv v
t
− .= = = =
.
 
(b) 24.06 m/sxv = since 0.xa = 0 19 55 m/s.y y yv v a t= + = − . 2 2 31.0 m/s.x yv v v= + = 
EVALUATE: Note that the speed is considerably less than the algebraic sum of the x- and y-components of 
the velocity. 
 3.14. IDENTIFY: Knowing the maximum reached by the froghopper and its angle of takeoff, we want to find its 
takeoff speed and the horizontal distance it travels while in the air. 
SET UP: Use coordinates with the origin at the ground and y+ upward. 0,xa = 29 80 m/sya = − . . At the 
maximum height 0yv = . The constant-acceleration formulas 2 2
0 02 ( )y y yv v a y y= + − and 
21
0 0 2y yy y v t a t− = + apply. 
EXECUTE: (a) 2 2
0 02 ( )y y yv v a y y= + − gives 
2
0 02 ( ) 2( 9.80 m/s )(0.587 m) 3.39 m/s.y yv a y y= − − = − − = 0 0 0sinyv v θ= so 
0
0
0
3 39 m/s 4 00 m/s.
sin sin58 0
yv
v
θ
.= = = .
. °
 
(b) Use the vertical motion to find the time in the air. When the froghopper has returned to the ground, 
0 0y y− = . 21
0 0 2y yy y v t a t− = + gives 0
2
2 2(3.39 m/s) 0.692 s.
9.80 m/s
y
y
v
t
a
= − = − =
−
 
Then 21
0 0 0 02 ( cos ) (4.00 m/s)(cos 58.0 )(0.692 s) 1.47 m.x xx x v t a t v tθ− = + = = ° = 
EVALUATE: 0yv = when 0
2
3 39 m/s 0 346 s
9 80 m/s
y
y
v
t
a
.= − = − = . .
− .
 The total time in the air is twice this. 
 3.15. IDENTIFY: The ball moves with projectile motion with an initial velocity that is horizontal and has 
magnitude 0.v The height h of the table and 0v are the same; the acceleration due to gravity changes from 
2
E 9 80 m/sg = . on earth to Xg on planet X. 
SET UP: Let x+ be horizontal and in the direction of the initial velocity of the marble and let y+ be 
upward. 0 0 ,xv v= 0 0,yv = 0,xa = ,ya g= − where g is either Eg or X.g 
EXECUTE: Use the vertical motion to find the time in the air: 0 .y y h− = − 21
0 0 2y yy y v t a t− = + gives 
2 .ht
g
= Then 21
0 0 2x xx x v t a t− = + gives 0 0 0
2 .x
hx x v t v
g
− = = 0x x D− = on earth and 2.76D on 
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Planet X. 0 0( ) 2 ,x x g v h− = which is constant, so E X2 76 .D g D g= . 
2E
X E2 0 131 1 28 m/s .
(2 76)
gg g= = . = .
.
 
EVALUATE: On Planet X the acceleration due to gravity is less, it takes the ball longer to reach the floor 
and it travels farther horizontally. 
 3.16. IDENTIFY: The shell moves in projectile motion. 
SET UP: Let x+ be horizontal, along the direction of the shell’s motion, and let y+ be upward. 0,xa = 
29.80 m/s .ya = − 
EXECUTE: (a) 0 0 0cos (40.0 m/s)cos 60.0 20.0 m/s,xv v α= = =° 
0 0 0sin (40.0 m/s)sin 60.0 34.6 m/s.yv v α= = =° 
(b) At the maximum height 0.yv = 0y y yv v a t= + gives 0
2
0 34.6 m/s 3.53 s.
9.80 m/s
y y
y
v v
t
a
− −= = =
−
 
(c) 2 2
0 02 ( )y y yv v a y y= + − gives 
2 2 2
0
0 2
0 (34.6 m/s) 61.2 m.
2 2( 9.80 m/s )
y y
y
v v
y y
a
− −− = = =
−
 
(d) The total time in the air is twice the time to the maximum height, so 
21
0 0 2 (20.0 m/s)(2)(3.53 s) 141 m.x xx x v t a t− = + = = 
(e) At the maximum height, 0 20.0 m/sx xv v= = and 0.yv = At all points in the motion, 0xa = and 
29.80 m/s .ya = − 
EVALUATE: The equation for the horizontal range R derived in the text is 
2
0 0sin 2 .vR
g
α= This gives 
2
2
(40.0 m/s) sin(120.0 ) 141 m,
9.80 m/s
R = =° which agrees with our result in part (d). 
 3.17. IDENTIFY: The baseball moves in projectile motion. In part (c) first calculate the components of the 
velocity at this point and then get the resultant velocity from its components. 
SET UP: First find the x- and y-components of the initial velocity. Use coordinates where the 
-directiony+ is upward, the -directionx+ is to the right and the origin is at the point where the baseball 
leaves the bat. 
 
0 0 0cos (30 0 m/s) cos36 9 24 0 m/sxv v α= = . . ° = . 
0 0 0sin (30 0 m/s) sin36 9 18 0 m/syv v α= = . . ° = . 
Figure 3.17a 
 
Use constant acceleration equations for the x and y motions, with 0xa = and ya g= − . 
EXECUTE: (a) y-component (vertical motion): 
0 10 0 m,y y− = + . 0 18 0 m/s,yv = . 29 80 m/s ,ya = − . ?t = 
21
0 0 2y yy y v a t− = + 
2 210.0 m (18.0 m/s) (4.90 m/s )t t= − 
2 2(4.90 m/s ) (18.0 m/s) 10.0 m 0t t− + = 
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Apply the quadratic formula: 21
9.80 18 0 ( 18 0) 4 (4 90)(10 0) s (1 837 1 154) st ⎡ ⎤= . ± − . − . . = . ± .⎢ ⎥⎣ ⎦
 
The ball is at a height of 10.0 above the point where it left the bat at 1 0 683 st = . and at 2 2 99 st = . . At the 
earlier time the ball passes through a height of 10.0 m as its way up and at the later time it passes through 
10.0 m on its way down. 
(b) 0 24 0 m/s,x xv v= = + . at all times since 0xa = . 
0y y yv v a t= + 
1 0 683 s:t = . 218.0 m/s ( 9.80 m/s )(0.683 s) 11.3 m/s.yv = + + − = + ( yv is positive means that the ball is 
traveling upward at this point.) 
2 2 99 s:t = . 218.0 m/s ( 9.80 m/s )(2.99 s) 11.3 m/s.yv = + + − = − ( yv is negative means that the ball is 
traveling downward at this point.) 
 (c) 0 24 0 m/sx xv v= = . 
Solve for :yv 
?,yv = 0 0y y− = (when ball returns to height where motion started), 
29 80 m/s ,ya = − . 0 18 0 m/syv = + . 
2 2
0 02 ( )y y yv v a y y= + − 
0 18 0 m/sy yv v= − = − . (negative, since the baseball must be traveling downward at this point) 
Now solve for the magnitude and direction of .Gv 
 
2 2
x yv v v= + 
2 2(24 0 m/s) ( 18 0 m/s) 30 0 m/sv = . + − . = .
18 0 m/stan
24 0 m/s
y
x
v
v
α − .= =
.
 
36.9 ,α = − ° 36 9. ° below the horizontal 
Figure 3.17b 
 
The velocity of the ball when it returns to the level where it left the bat has magnitude 30.0 m/s and is 
directed at an angle of 36 9. ° below the horizontal. 
EVALUATE: The discussion in parts (a) and (b) explains the significance of two values of t for which 
0 10 0 my y− = + . . When the ball returns to its initial height, our results give that its speed is the same as its 
initial speed and the angle of its velocity below the horizontal is equal to the angle of its initial velocity 
above the horizontal; both of these are general results. 
 3.18. IDENTIFY: The shot moves in projectile motion. 
SET UP: Let y+ be upward. 
EXECUTE: (a) If air resistance is to be ignored, the components of acceleration are 0 horizontally and 
29 80 m/sg− = − . vertically downward. 
(b) The x-component of velocity is constant at (12 0 m/s)cos51 0 7 55 m/s.xv = . . ° = . The y-component is 
0 (12 0 m/s) sin51 0 9 32 m/syv = . . ° = . at release and 
0 (9.32 m/s) (9.80 m/s)(2.08 s) 11.06 m/sy yv v gt= − = − = − when the shot hits. 
(c) 0 0 (7.55 m/s)(2.08 s) 15.7 m.xx x v t− = = = 
(d) The initial and final heights are not the same. 
(e) With 0y = and 0 yv as found above, the equation for y – y0 as a function of time gives 0 1 81m.y = . 
(f) The graphs are sketched in Figure 3.18. 
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EVALUATE: When the shot returns to its initial height, 9 32 m/s.yv = − . The shot continues to accelerate 
downward as it travels downward 1.81 m to the ground and the magnitude of yv at the ground is larger 
than 9.32 m/s. 
 
 
Figure 3.18 
 
 3.19. IDENTIFY: Take the origin of coordinates at the point where the quarter leaves your hand and take 
positive y to be upward. The quarter moves in projectile motion, with 0,xa = and ya g= − . It travels 
vertically for the time it takes it to travel horizontally 2.1 m. 
 
0 0 0cos (6.4 m/s) cos60xv v α= = ° 
0 3.20 m/sxv = 
0 0 0sin (6.4 m/s) sin 60yv v α= = ° 
0 5.54 m/syv = 
Figure 3.19 
 
(a) SET UP: Use the horizontal (x-component) of motion to solve for t, the time the quarter travels 
through the air: 
?,t = 0 2 1 m,x x− = . 0 3 2 m/s,xv = . 0xa = 
21
0 0 02 ,x x xx x v t a t v t− = + = since 0xa = 
EXECUTE: 0
0
2 1 m 0 656 s
3 2 m/sx
x xt
v
− .= = = .
.
 
SET UP: Now find the vertical displacement of the quarter after this time: 
0 ?,y y− = 29 80 m/s ,ya = − . 0 5 54 m/s,yv = + . 0 656 st = . 
21
0 0 2y yy y v t a t− + + 
EXECUTE: 2 21
0 2(5.54 m/s)(0.656 s) ( 9.80 m/s )(0.656 s) 3.63 m 2.11 m 1.5 m.y y− = + − = − = 
(b) SET UP: ?,yv = 0 656 s,t = . 29 80 m/s ,ya = − . 0 5 54 m/syv = + . 0y y yv v a t= + 
EXECUTE: 25.54 m/s ( 9.80 m/s )(0.656 s) 0.89 m/s.yv = + − = − 
EVALUATE: The minus sign for yv indicates that the y-component of Gv is downward. At this point the 
quarter has passed through the highest point in its path and is on its way down. The horizontal range if it 
returned to its original height (it doesn’t!) would be 3.6 m. It reaches its maximum height after traveling 
horizontally 1.8 m, so at 0 2 1 mx x− = . it is on its way down. 
 3.20. IDENTIFY: Consider the horizontal and vertical components of the projectile motion. The water travels 
45.0 m horizontally in 3.00 s. 
SET UP: Let y+ be upward. 0,xa = 29 80 m/s .ya = − . 0 0 0cos ,xv v θ= 0 0 0sin .yv v θ= 
EXECUTE: (a) 21
0 0 2x xx x v t a t− = + gives 0 0 0(cos )x x v tθ− = and 0
45.0 mcos 0.600;
(25.0 m/s)(3.00 s)
θ = =
 
 
0 53.1θ = ° 
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(b) At the highest point 0 (25.0 m/s)cos 53.1 15.0 m/s,x xv v= = ° = 0yv= and 2 2 15 0 m/s.x yv v v= + = . At 
all points in the motion, 29 80 m/sa = . downward. 
(c) Find 0y y− when 3 00s:t = . 
2 2 21 1
0 0 2 2(25.0 m/s)(sin53.1 )(3.00 s) ( 9.80 m/s )(3.00 s) 15.9 my yy y v t a t− = + = ° + − = 
0 15 0 m/s,x xv v= = . 2
0 (25 0 m/s)(sin53 1 ) (9 80m/s )(3 00 s) 9 41 m/s,y y yv v a t= + = . . ° − . . = − . and 
2 2 2 2(15 0 m/s) ( 9 41 m/s) 17 7 m/sx yv v v= + = . + − . = . 
EVALUATE: The acceleration is the same at all points of the motion. It takes the water 
0
2
20 0 m/s 2 04 s
9 80 m/s
y
y
v
t
a
.= − = − = .
− .
 to reach its maximum height. When the water reaches the building it has 
passed its maximum height and its vertical component of velocity is downward. 
 3.21. IDENTIFY: Take the origin of coordinates at the roof and let the -directiony+ be upward. The rock moves 
in projectile motion, with 0xa = and ya g= − . Apply constant acceleration equations for the x and y 
components of the motion. 
SET UP: 
 
0 0 0cos 25 2 m/sxv v α= = . 
0 0 0sin 16 3 m/syv v α= = . 
Figure 3.21a 
 
(a) At the maximum height 0yv = . 
29 80 m/s ,ya = − . 0,yv = 0 16 3 m/s,yv = + . 0 ?y y− = 
2 2
0 02 ( )y y yv v a y y= + − 
EXECUTE: 
2 2 2
0
0 2
0 (16.3 m/s) 13.6 m
2 2( 9.80 m/s )
y y
y
v v
y y
a
− −− = = = +
−
 
(b) SET UP: Find the velocity by solving for its x and y components. 
0 25 2 m/sx xv v= = . (since 0)xa = 
?,yv = 29 80 m/s ,ya = − . 0 15 0 my y− = − . (negative because at the ground the rock is below its initial 
position), 0 16 3 m/syv = . 
2 2
0 02 ( )y y yv v a y y= + − 
2
0 02 ( )y y yv v a y y= − + − ( yv is negative because at the ground the rock is traveling downward.) 
EXECUTE: 2 2(16.3 m/s) 2( 9.80 m/s )( 15.0 m) 23.7 m/syv = − + − − = − 
Then 2 2 2 2(25 2 m/s) ( 23 7 m/s) 34 6 m/sx yv v v= + = . + − . = . . 
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(c) SET UP: Use the vertical motion (y-component) to find the time the rock is in the air: 
?,t = 23 7 m/syv = − . (from part (b)), 29 80 m/s ,ya = − . 0 16 3 m/syv = + . 
EXECUTE: 0
2
23 7 m/s 16 3 m/s 4 08 s
9 80 m/s
y y
y
v v
t
a
− − . − .= = = + .
− .
 
SET UP: Can use this t to calculate the horizontal range: 
4 08 s,t = . 0 25 2 m/s,xv = . 0,xa = 0 ?x x− = 
EXECUTE: 21
0 0 2 (25 2 m/s)(4 08 s) 0 103 mx xx x v t a t− = + = . . + = 
(d) Graphs of x versus t, y versus t, xv versus t and yv versus t: 
 
 
Figure 3.21b 
 
EVALUATE: The time it takes the rock to travel vertically to the ground is the time it has to travel 
horizontally. With 0 16 3 m/syv = + . the time it takes the rock to return to the level of the roof ( 0)y = is 
02 / 3.33 s.yt v g= = The time in the air is greater than this because the rock travels an additional 15.0 m to 
the ground. 
 3.22. IDENTIFY and SET UP: The stone moves in projectile motion. Its initial velocity is the same as that of the 
balloon. Use constant acceleration equations for the x and y components of its motion. Take y+ to be 
downward. 
EXECUTE: (a) Use the vertical motion of the rock to find the initial height. 
5 00 s,t = . 0 20 0 m/s,yv = + . 29 80 m/s ,ya = + . 0 ?y y− = 
21
0 0 2y yy y v t a t− = + gives 0 223 my y− = . 
(b) In 5.00 s the balloon travels downward a distance 0 (20.0 m/s)(5.00 s) 100 m.y y− = = So, its height 
above ground when the rock hits is 223 m 100 m 123 m− = . 
(c) The horizontal distance the rock travels in 5.00 s is (15.0 m/s)(5.00 s) = 75.0 m. The vertical component 
of the distance between the rock and the basket is 123 m, so the rock is 2 2(75 m) (123 m) 144 m+ = 
from the basket when it hits the ground. 
(d) (i) The basket has no horizontal velocity, so the rock has horizontal velocity 15.0 m/s relative to the 
basket. Just before the rock hits the ground, its vertical component of velocity is 
2
0 20.0 m/s (9.80 m/s )(5.00 s) 69.0 m/s,y y yv v a t= + = + = downward, relative to the ground. The basket is 
moving downward at 20.0 m/s, so relative to the basket the rock has a downward component of velocity 49.0 m/s. 
(ii) horizontal: 15.0 m/s; vertical: 69.0 m/s 
EVALUATE: The rock has a constant horizontal velocity and accelerates downward. 
 3.23. IDENTIFY: Circular motion. 
SET UP: Apply the equation arad = 4π2R/T2, where T = 24 h. 
EXECUTE: (a) 
[ ]
2 6
2 3
rad 2
4 (6.38 10 m) 0.034 m/s 3.4 10 .
(24 h)(3600 s/h)
a gπ −×= = = × 
(b) Solving the equation arad = 4π2R/T2 for the period T with rad ,a g= 
2 6
2
4 (6.38 10 m) 5070 s 1.4 h.
9.80 m/s
T π ×= = = 
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EVALUATE: rada is proportional to 21 ,/T so to increase rada by a factor of 3
1 294
3 4 10− =
. ×
 requires 
that T be multiplied by a factor of 1 .
294
24 h 1 4 h.
294
= . 
 3.24. IDENTIFY: We want to find the acceleration of the inner ear of a dancer, knowing the rate at which she spins. 
SET UP: 0.070 m.R = For 3.0 rev/s, the period T (time for one revolution) is 1.0 s 0.333 s.
3.0 rev
T = = The 
speed is v = d/T = (2πR)/T, and 2
rad / .a v R= 
EXECUTE: 
2 2 2 2
2
rad 2 2
(2 / ) 4 4 (0.070 m) 25 m/s 2.5 .
(0.333 s)
v R T Ra g
R R T
π π π= = = = = = 
EVALUATE: The acceleration is large and the force on the fluid must be 2.5 times its weight. 
 3.25. IDENTIFY: For the curved lowest part of the dive, the pilot’s motion is approximately circular. We know 
the pilot’s acceleration and the radius of curvature, and from this we want to find the pilot’s speed. 
SET UP: 2
rad 5.5 53.9 m/s .a g= = 1 mph 0.4470 m/s.=
2
rad .va
R
= 
EXECUTE: 
2
rad ,va
R
= so 2
rad (280 m)(53.9 m/s ) 122.8 m/s 274.8 mph.v Ra= = = = Rounding these 
answers to 2 significant figures (because of 5.5g), gives v = 120 m/s = 270 mph. 
EVALUATE: This speed is reasonable for the type of plane flown by a test pilot. 
 3.26. IDENTIFY: Each blade tip moves in a circle of radius 3 40 mR = . and therefore has radial acceleration 
2
rad / .a v R= 
SET UP: 550 rev/min 9 17 rev/s,= . corresponding to a period of 1 0 109 s.
9 17 rev/s
T = = .
.
 
EXECUTE: (a) 2 196 m/s.Rv
T
π= = 
(b) 
2
4 2 3
rad 1 13 10 m/s 1 15 10 .va g
R
= = . × = . × 
EVALUATE: 
2
rad 2
4 Ra
T
π= gives the same results for rada as in part (b). 
 3.27. IDENTIFY: Uniform circular motion. 
SET UP: Since the magnitude of Gv is constant, tan 0
d
v
dt
= =
Gv
 and the resultant acceleration is equal to 
the radial component. At each point in the motion the radial component of the acceleration is directed in 
toward the center of the circular path and its magnitude is given by 2/ .v R 
EXECUTE: (a) 
2 2
2
rad
(6 00 m/s) 2.57 m/s ,
14 0 m
va
R
.= = =
.
 upward. 
(b) The radial acceleration has the same magnitude as in part (a), but now the direction toward the center of 
the circle is downward. The acceleration at this point in the motion is 22.57 m/s , downward. 
(c) SET UP: The time to make one rotation is the period T, and the speed v is the distance for one 
revolution divided by T. 
EXECUTE: 2 Rv
T
π= so 2 2 (14 0 m) 14.7 s.
6 00 m/s
RT
v
π π .= = =
.
 
EVALUATE: The radial acceleration is constant in magnitude since v is constant and is at every point in 
the motion directed toward the center of the circular path. The acceleration is perpendicular to Gv and is 
nonzero because the direction of Gv changes. 
 3.28. IDENTIFY: Each planet moves in a circular orbit and therefore has acceleration 2
rad .a v /R= 
Motion in Twoor Three Dimensions 3-15 
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SET UP: The radius of the earth’s orbit is 111 50 10 mr = . × and its orbital period is 
7365 days 3 16 10 s.T = = . × For Mercury, 105 79 10 mr = . × and 688 0 days 7 60 10 s.T = . = . × 
EXECUTE: (a) 42 2 98 10 m/srv
T
π= = . × 
(b) 
2
3 2
rad 5 91 10 m/s .va
r
−= = . × 
(c) 44 79 10 m/s,v = . × and 2 2
rad 3 96 10 m/s .a −= . × 
EVALUATE: Mercury has a larger orbital velocity and a larger radial acceleration than earth. 
 3.29. IDENTIFY: Each part of his body moves in uniform circular motion, with 
2
rad .va
R
= The speed in rev/s is 
1/ ,T where T is the period in seconds (time for 1 revolution). The speed v increases with R along the 
length of his body but all of him rotates with the same period T. 
SET UP: For his head 8 84 mR = . and for his feet 6 84 m.R = . 
EXECUTE: (a) 2
rad (8.84 m)(12.5)(9.80 m/s ) 32.9 m/sv Ra= = = 
(b) Use 
2
rad 2
4 .Ra
T
π= Since his head has rad 12 5a g= . and 8 84 m,R = . 
2
rad
8.84m2 2 1.688s.
12.5(9.80m/s )
RT
a
π π= = = Then his feet have 
2
2
rad 2 2
4 (6.84m) 94.8m/s
(1.688s)
Ra
T
π= = = = 9.67 g. 
The difference between the acceleration of his head and his feet is 212 5 9 67 2 83 27 7 m/s .g g g. − . = . = . 
(c) 1 1 0 592 rev/s 35 5 rpm
1 69 sT
= = . = .
.
 
EVALUATE: His feet have speed 2
rad (6 84 m)(94 8 m/s ) 25 5 m/s.v Ra= = . . = . 
 3.30. IDENTIFY: The relative velocities are S/F,Gv the velocity of the scooter relative to the flatcar, S/G ,Gv the 
scooter relative to the ground and F/G ,Gv the flatcar relative to the ground. S/G S/F F/G.= +G G Gv v v Carry out the 
vector addition by drawing a vector addition diagram. 
SET UP: S/F S/G F/G .= −G G Gv v v F/G
Gv is to the right, so F/G−Gv is to the left. 
EXECUTE: In each case the vector addition diagram gives 
(a) 5 0 m/s. to the right 
(b) 16.0 m/s to the left 
(c) 13 0 m/s. to the left. 
EVALUATE: The scooter has the largest speed relative to the ground when it is moving to the right relative 
to the flatcar, since in that case the two velocities S/F
Gv and F/G
Gv are in the same direction and their 
magnitudes add. 
 3.31. IDENTIFY: Relative velocity problem. The time to walk the length of the moving sidewalk is the length 
divided by the velocity of the woman relative to the ground. 
SET UP: Let W stand for the woman, G for the ground and S for the sidewalk. Take the positive direction 
to be the direction in which the sidewalk is moving. 
The velocities are W/Gv (woman relative to the ground), W/Sv (woman relative to the sidewalk), and S/Gv 
(sidewalk relative to the ground). 
The equation for relative velocity becomes W/G W/S S/Gv v v= + . 
The time to reach the other end is given by 
W/G
distance traveled relative to groundt
v
= 
EXECUTE: (a) S/G 1 0 m/sv = . 
W/S 1 5 m/sv = + . 
W/G W/S S/G 1 5 m/s 1 0 m/s 2 5 m/sv v v= + = . + . = . . 
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W/G
35 0 m 35 0 m 14 s
2 5 m/s
t
v
. .= = = .
.
 
 (b) S/G 1 0 m/sv = . 
W/S 1 5 m/sv = − . 
W/G W/S S/G 1 5 m/s 1 0 m/s 0 5 m/sv v v= + = − . + . = − . . (Since W/Gv now is negative, she must get on the 
moving sidewalk at the opposite end from in part (a).) 
W/G
35 0 m 35 0 m 70 s
0 5 m/s
t
v
− . − .= = = .
− .
 
EVALUATE: Her speed relative to the ground is much greater in part (a) when she walks with the motion 
of the sidewalk. 
 3.32. IDENTIFY: Calculate the rower’s speed relative to the shore for each segment of the round trip. 
SET UP: The boat’s speed relative to the shore is 6.8 km/h downstream and 1.2 km/h upstream. 
EXECUTE: The walker moves a total distance of 3.0 km at a speed of 4.0 km/h, and takes a time of three 
fourths of an hour (45.0 min). 
The total time the rower takes is 1 5 km 1 5 km 1 47 h 88 2 min
6 8 km/h 1 2 km/h
. .+ = . = . .
. .
 
EVALUATE: It takes the rower longer, even though for half the distance his speed is greater than 4.0 km/h. 
The rower spends more time at the slower speed. 
 3.33. IDENTIFY: Apply the relative velocity relation. 
SET UP: The relative velocities are C/E ,Gv the canoe relative to the earth, R/E ,Gv the velocity of the river 
relative to the earth and C/R ,Gv the velocity of the canoe relative to the river. 
EXECUTE: C/E C/R R/E= +G G Gv v v and therefore C/R C/E R/E.= −G G Gv v v The velocity components of C/R
Gv are 
0.50 m/s (0.40 m/s)/ 2, east and (0.40 m/s)/ 2, south,− + for a velocity relative to the river of 0.36 m/s, 
at 52 5. ° south of west. 
EVALUATE: The velocity of the canoe relative to the river has a smaller magnitude than the velocity of 
the canoe relative to the earth. 
 3.34. IDENTIFY: Relative velocity problem in two dimensions. 
(a) SET UP: P/A
Gv is the velocity of the plane relative to the air. The problem states that P A
Gv has 
magnitude 35 m/s and direction south. 
A/E
Gv is the velocity of the air relative to the earth. The problem states that A/E
Gv is to the southwest 
( 45 S° of W) and has magnitude 10 m/s. 
The relative velocity equation is P/E P/A A/E= + .G G Gv v v 
 
 
Figure 3.34a 
 
EXECUTE: (b) P/A( ) 0,xv = P/A( ) 35 m/syv = − 
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A/E( ) (10 m/s)cos 45 7.07 m/s,xv = − ° = − 
A/E( ) (10 m/s)sin 45 7.07 m/syv = − ° = − 
P/E P/A A/E( ) ( ) ( ) 0 7 07 m/s 7 1 m/sx x xv v v= + = − . = − . 
P/E P/A A/E( ) ( ) ( ) 35 m/s 7 07 m/s 42 m/sy y yv v v= + = − − . = − 
 
(c) 
 
2 2
P/E P/E P/E( ) ( )x yv v v= + 
2 2
P/E ( 7 1 m/s) ( 42 m/s) 43 m/sv = − . + − =
P/E
P/E
( ) 7 1tan 0 169
( ) 42
x
y
v
v
φ − .= = = .
−
 
9 6 ;φ = . ° ( 9 6. ° west of south) 
Figure 3.34b 
 
 
 
EVALUATE: The relative velocity addition diagram does not form a right triangle so the vector addition 
must be done using components. The wind adds both southward and westward components to the velocity 
of the plane relative to the ground. 
 3.35. IDENTIFY: Relative velocity problem in two dimensions. His motion relative to the earth (time 
displacement) depends on his velocity relative to the earth so we must solve for this velocity. 
(a) SET UP: View the motion from above. 
 
 The velocity vectors in the problem are: 
M/E ,Gv the velocity of the man relative to the earth 
W/E ,Gv the velocity of the water relative to the earth 
M/W ,Gv the velocity of the man relative to the water 
The rule for adding these velocities is 
M/E M/W W/E= +G G Gv v v 
Figure 3.35a 
The problem tells us that W/E
Gv has magnitude 2.0 m/s and direction due south. It also tells us that M/W
Gv 
has magnitude 4.2 m/s and direction due east. The vector addition diagram is then as shown in Figure 3.35b. 
 
 
This diagram shows the vector addition 
M/E M/W W/E= +G G Gv v v 
and also has M/W
Gv and W/E
Gv in their 
specified directions. Note that the vector 
diagram forms a right triangle. 
Figure 3.35b 
 
The Pythagorean theorem applied to the vector addition diagram gives 2 2 2
M/E M/W W/Ev v v= + . 
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EXECUTE: 2 2 2 2
M/E M/W W/E (4 2 m/s) (2 0 m/s) 4 7 m/s;v v v= + = . + . = . M/W
W/E
4 2 m/stan 2 10;
2 0 m/s
v
v
θ .= = = .
.
 
65 ;θ = ° or 90 25φ θ= ° − = °. The velocity of the man relative to the earth has magnitude 4.7 m/s and 
direction 25 S° of E. 
(b) This requires careful thought. To cross the river the man must travel 500 m due east relative to the 
earth. The man’s velocity relative to the earth is M/E .Gv But, from the vector addition diagram the eastward 
component of M/Ev equals M/W 4 2 m/sv = . . 
Thus 0 500 m 119 s,
4 2 m/sx
x xt
v
−= = =
.
which we round to 120 s. 
(c) The southward component of M/E
Gv equals W/E 2 0 m/sv = . . Therefore, in the 120 s it takes him to cross 
the river, the distance south the man travels relative to the earth is 
0 (2 0 m/s)(119 s) 240 myy y v t− = = . = . 
EVALUATE: If there were no current he would cross in the same time, (500 m)/(4 2 m/s) 120 s. = . The 
current carries him downstream but doesn’t affect his motion in the perpendicular direction, from bank to bank. 
 3.36. IDENTIFY: Use the relation that relates the relative velocities. 
SET UP: The relative velocities are the water relative to the earth, W/E ,Gv the boat relative to the water, 
B/W ,Gv and the boat relative to the earth, B/E.Gv B/E
Gv is due east, W/E
Gv is due south and has magnitude 
2.0 m/s. B/W 4 2 m/s.v = . B/E B/W W/E.= +G G Gv v v The velocity addition diagram is given in Figure 3.36. 
EXECUTE: (a) Find the direction of B/W.Gv W/E
B/W
2 0 m/ssin .
4 2 m/s
v
v
θ .= =
.
 28 4 ,θ = . ° north of east. 
(b) 2 2 2 2
B/E B/W W/E (4 2 m/s) (2 0 m/s) 3 7 m/sv v v= − = . − . = . 
(c) 
B/E
800 m 800 m 216 s.
3 7 m/s
t
v
= = =
.
 
EVALUATE: It takes longer to cross the river in this problem than it did in Problem 3.35. In the direction 
straight across the river (east) the component of his velocity relative to the earth is lass than 4.2 m/s. 
 
 
Figure 3.36 
 3.37. IDENTIFY: The resultant velocity, relative to the ground, is directly southward. This velocity is the sum of 
the velocity of the bird relative to the air and the velocity of the air relative to the ground. 
SET UP: B/A 100 km/h.=v A/G 40 km/h, east.=Gv B/G B/A A/G.= +G G Gv v v 
EXECUTE: We want B/G
Gv to be due south. The relative velocity addition diagram is shown in 
Figure 3.37. 
 
Motion in Two or Three Dimensions 3-19 
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Figure 3.37 
 
(a) A/G
B/A
40 km/hsin ,
100 km/h
v
v
φ = = 24 ,φ = ° west of south. 
(b) 2 2
B/G B/A A/G 91 7 km/hv v v= − = . . 
B/G
500 km 5 5 h.
91 7 km/h
dt
v
= = = .
.
 
EVALUATE: The speed of the bird relative to the ground is less than its speed relative to the air. Part of its 
velocity relative to the air is directed to oppose the effect of the wind. 
 3.38. IDENTIFY: Use the relation that relates the relative velocities. 
SET UP: The relative velocities are the velocity of the plane relative to the ground, P/G ,Gv the velocity of 
the plane relative to the air, P/A ,Gv and the velocity of the air relative to the ground, A/G.Gv P/G
Gv must be 
due west and A/G
Gv must be south. A/G 80 km/hv = and P/A 320 km/h.v = P/G P/A A/G .= +G G Gv v v The relative 
velocity addition diagram is given in Figure 3.38. 
EXECUTE: (a) A/G
P/A
80 km/hsin
320 km/h
v
v
θ = = and 14 ,θ = ° north of west. 
(b) 2 2 2 2
P/G P/A A/G (320 km/h) (80 0 km/h) 310 km/h.v v v= − = − . = 
EVALUATE: To travel due west the velocity of the plane relative to the air must have a westward 
component and also a component that is northward, opposite to the wind direction. 
 
 
Figure 3.38 
 
 3.39. IDENTIFY: d
dt
=
GG rv and d
dt
=
GG va 
SET UP: 1( ) .n nd t nt
dt
−= At 1 00 s,t = . 24 00 m/sxa = . and 23 00 m/s .ya = . At 0,t = 0x = and 
50 0 m.y = . 
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EXECUTE: (a) 2 .x
dxv Bt
dt
= = 2 ,x
x
dva B
dt
= = which is independent of t. 24 00 m/sxa = . gives 
22 00 m/s .B = . 23 .y
dyv Dt
dt
= = 6 .y
y
dv
a Dt
dt
= = 23 00 m/sya = . gives 30 500 m/s .D = . 0x = at 0t = 
gives 0.A = 50 0 my = . at 0t = gives 50 0 m.C = . 
(b) At 0,t = 0xv = and 0,yv = so 0.=Gv At 0,t = 22 4 00 m/sxa B= = . and 0,ya = so 
2 ˆ(4 00 m/s ) .= .Ga i 
(c) At 10 0 s,t = . 22 (2 00 m/s )(10 0 s) 40 0 m/sxv = . . = . and 3 23(0.500 m/s )(10.0 s) 150 m/s.yv = = 
2 2 155 m/s.x yv v v= + = 
(d) 2 2(2 00 m/s )(10 0 s) 200 m,x = . . = 3 350 0 m (0 500 m/s )(10 0 s) 550 m.y = . + . . = 
ˆ ˆ(200 m) (550 m) .= +Gr i j 
EVALUATE: The velocity and acceleration vectors as functions of time are 
2ˆ ˆ( ) (2 ) (3 )t Bt Dt= +Gv i j and ˆ ˆ( ) (2 ) (6 ) .t B Dt= +Ga i j The acceleration is not constant. 
 
 3.40. IDENTIFY: The acceleration is not constant but is known as a function of time. 
SET UP: Integrate the acceleration to get the velocity and the velocity to get the position. At the maximum 
height 0.yv = 
EXECUTE: (a) 3 2
0 0, ,
3 2x x y yv v t v v t tα γβ= + = + − and 4 2 3
0 0, .
12 2 6x yx v t t y v t t tα β γ= + = + − 
(b) Setting 0yv = yields a quadratic in 2
0, 0 .
2yt v t tγβ= + − Using the numerical values given in the 
problem, this equation has as the positive solution 2
0
1 2 13 59 s.yt vβ β γ
γ
⎡ ⎤= + + = .⎢ ⎥⎣ ⎦
 Using this time in 
the expression for y(t) gives a maximum height of 341 m. 
(c) 0y = gives 2 3
00
2 6yv t t tβ γ= + − and 2
0 0.
6 2 yt t vγ β− − = Using the numbers given in the problem, the 
positive solution is t = 20.73 s. For this t, 43 85 10 m.x = . × 
EVALUATE: We cannot use the constant-acceleration kinematics formulas, but calculus provides the 
solution. 
 3.41. IDENTIFY: .d dtG Gv = r/ This vector will make a 45° angle with both axes when its x- and y-components 
are equal. 
 SET UP: 1( ) .
n
nd t nt
dt
−= 
 EXECUTE: 2ˆ ˆ2 3 .bt ctGv = i + j x yv v= gives 2 3 .t b c= 
 EVALUATE: Both components of Gv change with t. 
 3.42. IDENTIFY: Use the position vector of a dragonfly to determine information about its velocity vector and 
acceleration vector. 
SET UP: Use the definitions / ,xv dx dt= / ,yv dy dt= / ,x xa dv dt= and / .y ya dv dt= 
EXECUTE: (a) Taking derivatives of the position vector gives the components of the velocity vector: 
2( ) (0.180 m/s ) ,xv t t= 3 2( ) ( 0.0450 m/s ) .yv t t= − Use these components and the given direction: 
3 2
o
2
(0.0450 m/s )tan30.0 ,
(0.180 m/s )
t
t
= which gives 2 31 s.t = . 
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 (b) Taking derivatives of the velocity components gives the acceleration components: 
20 180 m/s ,xa = . 3( ) 0.0900 m/s ) .ya t t= −( At 2 31 s,t = . 20 180 m/sxa = . and 20 208 m/s ,ya = − . giving 
20 275 m/s .a = . The direction is 0 208tan ,
0 180
θ .=
.
 so o49 1θ = . clockwise from +x-axis. 
 EVALUATE: The acceleration is not constant, so we cannot use the standard kinematics formulas. 
 3.43. IDENTIFY: Once the rocket leaves the incline it moves in projectile motion. The acceleration along the 
incline determines the initial velocity and initial position for the projectile motion.SET UP: For motion along the incline let x+ be directed up the incline. 2 2
0 02 ( )x x xv v a x x= + − gives 
22(1.90 m/s )(200 m) 27.57 m/s.xv = = When the projectile motion begins the rocket has 0 27.57 m/sv = 
at 35 0. ° above the horizontal and is at a vertical height of (200 0 m) sin35 0 114 7 m.. . ° = . For the 
projectile motion let x+ be horizontal to the right and let y+ be upward. Let 0y = at the ground. Then 
0 114 7 m,y = . 0 0 cos35 0 22.57 m/s,xv v= . ° = 0 0 sin35 0 15.81 m/s,yv v= . ° = 0,xa = 29 80 m/s .ya = − . Let 
0x = at point A, so 0 (200 0 m)cos35 0 163 8 m.x = . . ° = . 
EXECUTE: (a) At the maximum height 0.yv = 2 2
0 02 ( )y y yv v a y y= + − gives 
2 2 2
0
0 2
0 (15 81 m/s) 12.77 m
2 2( 9 80 m/s )
y y
y
v v
y y
a
− − .− = = =
− .
 and 114 7 m 12.77 m 128 m.y = . + = The maximum height 
above ground is 128 m. 
(b) The time in the air can be calculated from the vertical component of the projectile motion: 
0 114.7 m,y y− =− 0 15.81 m/s,yv = 29 80 m/s .ya = − . 21
0 0 2y yy y v t a t− = + gives 
2 2(4.90 m/s ) (15.81 m/s) 114.7 m.t t− − The quadratic formula gives 6 713 st = . for the positive root. Then 
21
0 0 2 (22.57 m/s)(6.713 s) 151.6 mx xx x v t a t− = + = = and 163 8 m 151.6 m 315 m.x = . + = The horizontal 
range of the rocket is 315 m. 
 EVALUATE: The expressions for h and R derived in the range formula do not apply here. They are only 
for a projectile fired on level ground. 
 3.44. IDENTIFY: 0 0
( )
t
t dt= + ∫
G G Gr r v and .d
dt
=
Gva 
SET UP: At 0,t = 0 0x = and 0 0.y = 
EXECUTE: (a) Integrating, 3 2ˆ ˆ.
3 2
t t tβ γα⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Gr i j Differentiating, ˆ ˆ( 2 ) .tβ γ= − +Ga i j 
(b) The positive time at which 0x = is given by 2 3 .t α β= At this time, the y-coordinate is 
2
2
3
3 3(2.4 m/s)(4.0 m/s ) 9.0 m.
2 2 2(1.6 m/s )
y tγ αγ
β
= = = = 
EVALUATE: The acceleration is not constant. 
 3.45. IDENTIFY: Take y+ to be downward. Both objects have the same vertical motion, with 0yv and 
ya g= + . Use constant acceleration equations for the x and y components of the motion. 
SET UP: Use the vertical motion to find the time in the air: 
0 0,yv = 29 80 m/s ,ya = . 0 25 m,y y− = ?t = 
EXECUTE: 21
0 0 2y yy y v t a t− = + gives 2 259 s.t = . 
During this time the dart must travel 90 m, so the horizontal component of its velocity must be 
0
0
70 m 31 m/s.
2 259 sx
x xv
t
−= = =
.
 
EVALUATE: Both objects hit the ground at the same time. The dart hits the monkey for any muzzle 
velocity greater than 31 m/s. 
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 3.46. IDENTIFY: The velocity has a horizontal tangential component and a vertical component. The vertical 
component of acceleration is zero and the horizontal component is 
2
rad .xva
R
= 
SET UP: Let y+ be upward and x+ be in the direction of the tangential velocity at the instant we are 
considering. 
EXECUTE: (a) The bird’s tangential velocity can be found from 
circumference 2 (6 00 m) 7.54 m/s.
time of rotation 5 00 sxv π .= = =
.
 
Thus its velocity consists of the components 7 54 m/sxv = . and 3 00 m/s.yv = . The speed relative to the 
ground is then 2 2 8 11 m/s.x yv v v= + = . 
(b) The bird’s speed is constant, so its acceleration is strictly centripetal—entirely in the horizontal 
direction, toward the center of its spiral path—and has magnitude 
2 2
2
rad
(7 54 m/s) 9.48 m/s .
6 00 m
xva
r
.= = =
.
 
(c) Using the vertical and horizontal velocity components 1 3 00 m/stan 21 7 .
7 54 m/s
θ − .= = . °
.
 
EVALUATE: The angle between the bird’s velocity and the horizontal remains constant as the bird rises. 
 3.47. IDENTIFY: The cannister moves in projectile motion. Its initial velocity is horizontal. Apply constant 
acceleration equations for the x and y components of motion. 
SET UP: 
 
 
 
Take the origin of coordinates at the point 
where the cannister is released. Take +y to be 
upward. The initial velocity of the cannister is 
the velocity of the plane, 64.0 m/s in the 
+x-direction. 
Figure 3.47 
 
Use the vertical motion to find the time of fall: 
?,t = 0 0,yv = 29 80 m/s ,ya = − . 0 90 0 my y− = − . (When the cannister reaches the ground it is 90.0 m 
below the origin.) 
21
0 0 2y yy y v t a t− = + 
EXECUTE: Since 0 0,yv = 0
2
2( ) 2( 90.0 m) 4.286 s.
9.80 m/sy
y yt
a
− −= = =
−
 
SET UP: Then use the horizontal component of the motion to calculate how far the cannister falls in this 
time: 
0 ?,x x− = 0,xa − 0 64 0 m/sxv = . 
EXECUTE: 21
0 0 2 (64.0 m/s)(4.286 s) 0 274 m.x x v t at− = + = + = 
EVALUATE: The time it takes the cannister to fall 90.0 m, starting from rest, is the time it travels 
horizontally at constant speed. 
 3.48. IDENTIFY: The person moves in projectile motion. Her vertical motion determines her time in the air. 
SET UP: Take y+ upward. 0 15 0 m/s,xv = . 0 10 0 m/s,yv = + . 0,xa = 29 80 m/s .ya = − . 
EXECUTE: (a) Use the vertical motion to find the time in the air: 21
0 0 2y yy y v t a t− = + with 
0 30 0 my y− = − . gives 2 230 0 m (10 0 m/s) (4 90 m/s ) .t t− . = . − . The quadratic formula gives 
Motion in Two or Three Dimensions 3-23 
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( )21 10.0 ( 10.0) 4(4.9)( 30) s.
2(4.9)
t = + ± − − − The positive solution is 3 70 s.t = . During this time she 
travels a horizontal distance 21
0 0 2 (15.0 m/s)(3.70 s) 55.5 m.x xx x v t a t− = + = = She will land 55.5 m south 
of the point where she drops from the helicopter and this is where the mats should have been placed. 
(b) The x-t, y-t, xv -t and yv -t graphs are sketched in Figure 3.48. 
EVALUATE: If she had dropped from rest at a height of 30.0 m it would have taken her 
2
2(30 0 m) 2 47 s.
9 80 m/s
t .= = .
.
 She is in the air longer than this because she has an initial vertical component of 
velocity that is upward. 
 
 
Figure 3.48 
 
 3.49. IDENTIFY: The suitcase moves in projectile motion. The initial velocity of the suitcase equals the velocity 
of the airplane. 
SET UP: Take y+ to be upward. 0,xa = .ya g= − 
EXECUTE: Use the vertical motion to find the time it takes the suitcase to reach the ground: 
2
0 0 0 sin 23 , 9.80 m/s , 114 m, ?y yv v a y y t= ° = − − = − = 21
0 0 2 gives 9.60 s.y yy y v t a t t− = + = 
The distance the suitcase travels horizontally is 0 0 0( cos23.0 ) 795 m.xx x v v t− = = ° = 
EVALUATE: An object released from rest at a height of 114 m strikes the ground at 
02( ) 4.82 s.y yt
g
−= =
−
 The suitcase is in the air much longer than this since it initially has an upward 
component of velocity. 
 3.50. IDENTIFY: The shell moves as a projectile. To just clear the top of the cliff, the shell must have 
0 25.0 my y− = when it has 0 60.0 m.x x− = 
SET UP: Let y+ be upward. 0,xa = .ya g= − 0 0 cos 43 ,xv v= ° 0 0 sin 43 .yv v= ° 
EXECUTE: (a) horizontal motion: 0 0
0
60.0 m so .
( cos43 )xx x v t t
v
− = =
°
 
vertical motion: 2 2 21 1
0 0 02 2 gives 25.0 m ( sin 43.0 ) ( 9.80 m/s ) .y yy y v t a t v t t− = + = ° + − 
Solving these two simultaneous equations for 0v and t gives 0 32.6 m/sv = and 2.51 s.t = 
(b) yv when shell reaches cliff: 
2
0 (32.6 m/s) sin 43.0 (9.80 m/s )(2.51 s) 2.4 m/sy y yv v a t= + = ° − = − 
The shell is traveling downward when it reaches the cliff, so it lands right at the edge of the cliff. 
EVALUATE: The shell reaches its maximum height at 0 2.27 s,y
y
v
t
a
= − = whichconfirms that at 
2.51 st = it has passed its maximum height and is on its way down when it strikes the edge of the cliff. 
3.51. IDENTIFY: Find the horizontal distance a rocket moves if it has a non-constant horizontal acceleration but 
a constant vertical acceleration of g downward. 
SET UP: The vertical motion is g downward, so we can use the constant acceleration formulas for that 
component of the motion. We must use integration for the horizontal motion because the acceleration is not 
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constant. Solving for t in the kinematics formula for y gives 02( ) .
y
y yt
a
−= In the horizontal direction we 
must use 0 0
( ) ( )
t
x x xv t v a t dt= + ′ ′∫ and 0 0
( ) .
t
xx x v t dt− = ′ ′∫ 
EXECUTE: Use vertical motion to find t. 0
2
2( ) 2(30.0 m) 2.474 s.
9.80 m/sy
y yt
a
−= = = 
In the horizontal direction we have 
3 2 3 2
0 00
( ) ( ) (0.800 m/s ) 12.0 m/s (0.800 m/s ) .
t
x x x xv t v a t dt v t t= + ′ ′ = + = +∫ Integrating ( )xv t gives 
3 3
0 (12.0 m/s) (0.2667 m/s ) .x x t t− = + At 2 474 s,t = . 0 29 69 m 4 04 m 33 7 m.x x− = . + . = . 
EVALUATE: The vertical part of the motion is familiar projectile motion, but the horizontal part is not. 
 3.52. IDENTIFY: The equipment moves in projectile motion. The distance D is the horizontal range of the 
equipment plus the distance the ship moves while the equipment is in the air. 
SET UP: For the motion of the equipment take x+ to be to the right and y+ to be upward. Then 0,xa = 
29 80 m/s ,ya = − . 0 0 0cos 7 50 m/sxv v α= = . and 0 0 0sin 13 0 m/s.yv v α= = . When the equipment lands in 
the front of the ship, 0 8 75 m.y y− = − . 
EXECUTE: Use the vertical motion of the equipment to find its time in the air: 21
0 0 2y yy y v t a t− = + gives 
( )21 13.0 ( 13.0) 4(4.90)(8.75) s.
9.80
t = ± − + The positive root is 3 21 s.t = . The horizontal range of the 
equipment is 21
0 0 2 (7.50 m/s)(3.21 s) 24.1 m.x xx x v t a t− = + = = In 3.21 s the ship moves a horizontal 
distance (0.450 m/s)(3.21 s) 1.44 m,= so 24 1 m 1 44 m 25 5 m.D = . + . = . 
EVALUATE: The range equation 
2
0 0sin 2vR
g
α= cannot be used here because the starting and ending 
points of the projectile motion are at different heights. 
 3.53. IDENTIFY: Projectile motion problem. 
 
 Take the origin of coordinates at the point 
where the ball leaves the bat, and take +y to be 
upward. 
0 0 0cosxv v α= 
0 0 0sin ,yv v α= 
but we don’t know 0v . 
Figure 3.53 
 
Write down the equation for the horizontal displacement when the ball hits the ground and the 
corresponding equation for the vertical displacement. The time t is the same for both components, so this 
will give us two equations in two unknowns 0(v and t). 
(a) SET UP: y-component: 
29 80 m/s ,ya = − . 0 0 9 m,y y− = − . 0 0 sin 45yv v= ° 
21
0 0 2y yy y v t a t− = + 
EXECUTE: 2 21
0 20.9 m ( sin 45 ) ( 9.80 m/s )v t t− = ° + − 
SET UP: x-component: 
0,xa = 0 188 m,x x− = 0 0 cos45xv v= ° 
21
0 0 2x xx x v t a t− = + 
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EXECUTE: 0
0 0
188 m
cos45x
x xt
v v
−= =
°
 
Put the expression for t from the x-component motion into the y-component equation and solve for 0v . 
(Note that sin 45 cos45° = °. ) 
2
2
0
0 0
188 m 188 m0 9 m ( sin 45 ) (4 90 m/s )
cos45 cos45
v
v v
⎛ ⎞ ⎛ ⎞
− . = ° − .⎜ ⎟ ⎜ ⎟° °⎝ ⎠ ⎝ ⎠
 
2
2
0
188 m4 90 m/s 188 m 0 9 m 188 9 m
cos45v
⎛ ⎞
. = + . = .⎜ ⎟°⎝ ⎠
 
2 2
0 cos45 4 90 m/s ,
188 m 188 9 m
v ° .⎛ ⎞ =⎜ ⎟ .⎝ ⎠
 
2
0
188 m 4 90 m/s 42 8 m/s
cos45 188 9 m
v .⎛ ⎞= = .⎜ ⎟° .⎝ ⎠
 
(b) Use the horizontal motion to find the time it takes the ball to reach the fence: 
SET UP: x-component: 
0 116 m,x x− = 0xa ,= 0 0 cos45 (42 8 m/s) cos45 30 3 m/s,xv v= ° = . ° = . ?t = 
21
0 0 2x xx x v t a t− = + 
EXECUTE: 0
0
116 m 3 83 s
30 3 m/sx
x xt
v
−= = = .
.
 
SET UP: Find the vertical displacement of the ball at this t: 
y-component: 
0 ?,y y− = 29 80 m/s ,ya = − . 0 0 sin 45 30 3 m/s,yv v= ° = . 3 83 st = . 
21
0 0 2y yy y v t a t− = + 
EXECUTE: 2 21
0 2(30 3 s)(3 83 s) ( 9 80 m/s )(3 83 s)y y− = . . + − . . 
0 116 0 m 71 9 m 44 1 m,y y− = . − . = + . above the point where the ball was hit. The height of the ball above 
the ground is 44 1 m 0 90 m 45 0 m. + . = . . Its height then above the top of the fence is 
45 0 m 3 0 m 42 0 m. − . = . . 
EVALUATE: With 0 42 8 m/s,v = . 0 30 3 m/syv = . and it takes the ball 6.18 s to return to the height where 
it was hit and only slightly longer to reach a point 0.9 m below this height. 0(188 m)/( cos45 )t v= ° gives 
6 21 s,t = . which agrees with this estimate. The ball reaches its maximum height approximately 
(188 m)/2 94 m= from home plate, so at the fence the ball is not far past its maximum height of 47.6 m, 
so a height of 45.0 m at the fence is reasonable. 
 3.54. IDENTIFY: While the hay falls 150 m with an initial upward velocity and with a downward acceleration 
of g, it must travel a horizontal distance (the target variable) with constant horizontal velocity. 
SET UP: Use coordinates with y+ upward and x+ horizontal. The bale has initial velocity components 
0 0 0cos (75 m/s)cos55 43.0 m/sxv v α= = ° = and 0 0 0sin (75 m/s)sin55 61.4 m/s.yv v α= = ° = 0 150 my = 
and 0.y = The equation 21
0 0 2y yy y v t a t− = + applies to the vertical motion and a similar equation to the 
horizontal motion. 
EXECUTE: Use the vertical motion to find t: 21
0 0 2y yy y v t a t− = + gives 
2 2150 m (61.4 m/s) (4.90 m/s ) .t t− = − The quadratic formula gives 6 27 8 36 s.t = . ± . The physical value 
is the positive one, and 14 6 s.t = . Then 21
0 0 2 (43.0 m/s)(14.6 s) 630 m.x xx x v t a t− = + = = 
EVALUATE: If the airplane maintains constant velocity after it releases the bales, it will also travel 
horizontally 630 m during the time it takes the bales to fall to the ground, so the airplane will be directly 
over the impact spot when the bales land. 
 3.55. IDENTIFY: Two-dimensional projectile motion. 
3-26 Chapter 3 
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SET UP: Let +y be upward. 0,xa = 29 80 m/sya = − . . With 0 0 0,x y= = algebraic manipulation of the 
equations for the horizontal and vertical motion shows that x and y are related by 
 2
0 2 2
0 0
(tan )
2 cos
gy x xθ
υ θ
= − . 
0 60 0θ = . °. 8 00 my = . when 18 0 mx = . . 
EXECUTE: (a) Solving for v0 gives 
2
0 2
0 0
16 6 m/s.
2(cos )( tan )
gx
x y
υ
θ θ
= = .
−
 
(b) We find the horizontal and vertical velocity components: 
0 0 0cos 8 3 m/sx xυ υ υ θ= = = . . 
2 2
0 02 ( )y y ya y yυ υ= + − gives 
2 2 2
0 0 0( sin ) 2 ( ) (14 4 m/s) 2( 9 80 m/s )(8 00 m) 7 1 m/sy ya y yυ υ θ= − + − = − . + − . . = − . 
2 2 10 9 m/sx yυ υ υ= + = . . 7 1tan
8 3
y
x
υ
θ
υ
.= =
.
| |
| |
 and 40 5 ,θ = . ° below the horizontal. 
EVALUATE: We can check our calculated 0v . 
0
0
18 0 m 2 17 s
8 3 m/sx
x xt
υ
− .= = = . .
.
 
Then 2 2 21
0 0 2 (14 4 m/s)(2 17 s) (4 9 m/s )(2 17 s) 8 m,y yy y t a tυ− = + = . . − . . = which checks. 
 3.56. IDENTIFY: The water moves in projectile motion. 
SET UP: Let 0 0 0x y= = and take y+ to be positive. 0,xa = .ya g= − 
EXECUTE: The equationsof motions are 21
0 2( sin )y v t gtα= − and 0( cos ) .x v tα= When the water 
goes in the tank for the minimum velocity, 2y D= and 6 .x D= When the water goes in the tank for the 
maximum velocity, 2y D= and 7 .x D= In both cases, sin cos 2 2/α α= = . 
To reach the minimum distance: 0
26 ,
2
D v t= and 21
0 2
22 .
2
D v t gt= − Solving the first equation for t 
gives 
0
6 2 .Dt
v
= Substituting this into the second equation gives 
2
1
2
0
6 22 6 .DD D g
v
⎛ ⎞
= − ⎜ ⎟⎜ ⎟
⎝ ⎠
 Solving this 
for 0v gives 0 3 .v gD= 
To reach the maximum distance: 0
27 ,
2
D v t= and 21
0 2
22 .
2
D v t gt= − Solving the first equation for t 
gives 
0
7 2 .Dt
v
= Substituting this into the second equation gives
2
1
2
0
7 22 7 .DD D g
v
⎛ ⎞
= − ⎜ ⎟⎜ ⎟
⎝ ⎠
 Solving this 
for 0v gives 0 49 5 3 13 ,v gD/ gD= = . which, as expected, is larger than the previous result. 
EVALUATE: A launch speed of 0 6 2 45v gD gD= = . is required for a horizontal range of 6D. The 
minimum speed required is greater than this, because the water must be at a height of at least 2D when it 
reaches the front of the tank. 
 3.57. IDENTIFY: From the figure in the text, we can read off the maximum height and maximum horizontal 
distance reached by the grasshopper. Knowing its acceleration is g downward, we can find its initial speed 
and the height of the cliff (the target variables). 
SET UP: Use coordinates with the origin at the ground and y+ upward. 0,xa = 29 80 m/sya = − . . The 
constant-acceleration kinematics formulas 2 2
0 02 ( )y y yv v a y y= + − and 21
0 0 2x xx x v t a t− = + apply. 
Motion in Two or Three Dimensions 3-27 
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EXECUTE: (a) 0yv = when 0 0 0674 m.y y− = . 2 2
0 02 ( )y y yv v a y y= + − gives 
2
0 02 ( ) 2 ( 9 80 m/s )(0 0674 m) 1 15 m/s.y yv a y y= − − = − − . . = . 0 0 0sinyv v α= so 
0
0
0
1 15 m/s 1 50 m/s.
sin sin50 0
yv
v
α
.= = = .
. °
 
(b) Use the horizontal motion to find the time in the air. The grasshopper travels horizontally 
0 1 06 mx x− = . . 21
0 0 2x xx x v t a t− = + gives 0 0
0 0
1.10 s.
cos50.0x
x x x xt
v v
− −= = =
°
 Find the vertical 
displacement of the grasshopper at 1.10 s:t = 
2 2 21 1
0 0 2 2(1 15 m/s)(1 10 s) ( 9 80 m/s )(1 10 s) 4 66 m.y yy y v t a t− = + = . . + − . . = − . The height of the cliff is 
4.66 m. 
EVALUATE: The grasshopper’s maximum height (6.74 cm) is physically reasonable, so its takeoff speed 
of 1.50 m/s must also be reasonable. Note that the equation 
2
0 0sin 2vR
g
α= does not apply here since the 
launch point is not at the same level as the landing point. 
 3.58. IDENTIFY: To clear the bar the ball must have a height of 10.0 ft when it has a horizontal displacement of 
36.0 ft. The ball moves as a projectile. When 0v is very large, the ball reaches the goal posts in a very short 
time and the acceleration due to gravity causes negligible downward displacement. 
SET UP: 36 0 ft 10.97 m;. = 10 0 ft 3 048 m.. = . Let x+ be to the right and y+ be upward, so 0,xa = 
,ya g= − 0 0 0cosxv v α= and 0 0 0sin .yv v α= 
EXECUTE: (a) The ball cannot be aimed lower than directly at the bar. 0
10 0 fttan
36 0 ft
α .=
.
 and 0 15.5 .α = ° 
(b) 21
0 0 2x xx x v t a t− = + gives 0 0
0 0 0
.
cosx
x x x xt
v v α
− −= = Then 21
0 0 2y yy y v t a t− = + gives 
2 2
0 0 0
0 0 0 0 02 2 2 2
0 0 0 0 0 0
1 ( ) 1 ( )( sin ) ( ) tan .
cos 2 2cos cos
x x x x x xy y v g x x g
v v v
α α
α α α
⎛ ⎞− − −− = − = − −⎜ ⎟
⎝ ⎠
 
2
0
0
0 0 0 0
( ) 10.97 m 9 80 m/s 12.2 m/s
cos 2[( ) tan ( )] cos45 0 2[10.97 m 3 048 m]
x x gv
x x y yα α
− .= = =
− − − . ° − .
= 43.9 km/h. 
EVALUATE: With the 0v and 45° launch angle in part (b), the horizontal range of the ball is 
2
0 0sin 2vR
g
α= = 15.2 m = 49.9 ft. The ball reaches the highest point in its trajectory when 
0 /2,x x R− = which is 25 ft, so when it reaches the goal posts it is on its way down. 
 3.59. IDENTIFY: The snowball moves in projectile motion. In part (a) the vertical motion determines the time in 
the air. In part (c), find the height of the snowball above the ground after it has traveled horizontally 4.0 m. 
SET UP: Let +y be downward. 0,xa = 29 80 m/s .ya = + . 0 0 0cos 5 36 m/s,xv v θ= = . 
0 0 0sin 4 50 m/s.yv v θ= = . 
EXECUTE: (a) Use the vertical motion to find the time in the air: 21
0 0 2y yy y v t a t− = + with 
0 14 0 my y− = . gives 2 214 0 m (4 50 m/s) (4 9 m/s ) .t t. = . + . The quadratic formula gives 
( )21 4 50 (4 50) 4 (4 9)( 14 0) s.
2(4 9)
t = − . ± . − . − .
.
 The positive root is 1 29 s.t = . Then 
21
0 0 2 (5 36 m/s)(1 29 s) 6 91 m.x xx x v t a t− = + = . . = . 
3-28 Chapter 3 
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 
 
(b) The x-t, y-t, -xv t and -yv t graphs are sketched in Figure 3.59. 
(c) 21
0 0 2x xx x v t a t− = + gives 0
0
4 0 m 0 746 s.
5 36 m/sx
x xt
v
− .= = = .
.
 In this time the snowball travels downward 
a distance 21
0 0 2 6 08 my yy y v t a t− = + = . and is therefore 14 0 m 6 08 m 7 9 m. − . = . above the ground. The 
snowball passes well above the man and doesn’t hit him. 
EVALUATE: If the snowball had been released from rest at a height of 14.0 m it would have reached the 
ground in 2
2(14 0 m) 1 69 s.
9 80 m/s
t .= = .
.
 The snowball reaches the ground in a shorter time than this because of 
its initial downward component of velocity. 
 
 
 
Figure 3.59 
 
 
 3.60. IDENTIFY: The dog runs horizontally at constant velocity, and the ball is in two-dimensional projectile 
motion. The ball starts out traveling only horizontally. 
SET UP: Use coordinates with the origin at the boy and with +y downward. For the ball 
2
0 00, 8 50 m/s, 0 and 9 80 m/s .y x x ya aυ υ= = . = = . 
EXECUTE: (a) The dog must travel horizontally the same distance the ball travels horizontally, so the dog 
must have speed 8 50 m/s. . 
(b) Use the vertical motion of the ball to find its time in the air. 21
0 0 2y yy y t a tυ− = + gives 
0
2
2( ) 2(12 0 m) 1 56 s
9 80 m/sy
y yt
a
− .= = = .
.
. Then 21
0 0 2 (8 50 m/s)(1 56 s) 13 3 mx xx x t a tυ− = + = . . = . 
EVALUATE: The dog is about 40 ft from the tree, which is not unreasonable since the tree is nearly 40 ft 
high. 
 3.61. IDENTIFY: The dog runs horizontally at constant velocity, and the ball is in two-dimensional projectile 
motion. But this time the ball has an upward component to its initial velocity. 
SET UP: Use coordinates with the origin at the boy and with +y upward. The ball has 0 0 0cosxυ υ θ= = 
(8 50 m/s)cos60 0 4 25 m/s,. . = .D
0 0 0sin (8 50 m/s)sin 60 0 7 36 m/s,yυ υ θ= = . . = .D 0xa = and 
29 80 m/sya = − . . 
EXECUTE: (a) The dog must travel horizontally the same distance the ball travels horizontally, so the dog 
must have speed 4 25 m/s. . 
(b) Use the vertical motion of the ball to find its time in the air. 21
0 0 2y yy y t a tυ− = + gives 
2 212 0 m (7 36 m/s) (4 90 m/s )t t− . = . − . . The quadratic formula gives 0 751 1 74 st = . ± . . The negative 
value is not physical, so 2 49 st = . . Then 0x x− = 21
0 2 (4 25 m/s)(2 49 s) 10 6 m.x xt a tυ + = . . = . 
EVALUATE: The ball is in the air longer than when it is thrown horizontally (as we saw in the previous 
problem), but it doesn’t travel as far horizontally. The dog doesn’t have to run as far or as fast as when the 
ball is thrown horizontally. 
 3.62. IDENTIFY: The rock moves in projectile motion. 
SET UP: Let y+ be upward. 0,xa = .ya g= − Eqs. (3.21) and (3.22) give